KnowledgeBoat Logo
|
OPEN IN APP

Chapter 16

Trigonometrical Ratios

Class - 9 ML Aggarwal Understanding ICSE Mathematics

Exercise 16

Question 1(a)

From the figure (1) given below, find the values of:

(i) sin θ

(ii) cos θ

(iii) tan θ

(iv) cot θ

(v) sec θ

(vi) cosec θ

From the figure, find values of (i) sin θ (ii) cos θ (iii) tan θ (iv) cot θ (v) sec θ (vi) cosec θ. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In right-angled triangle OMP,

By Pythagoras theorem, we get

⇒ OP2 = OM2 + MP2

⇒ MP2 = OP2 - OM2

⇒ MP2 = (15)2 - (12)2

⇒ MP2 = 225 - 144

⇒ MP2 = 81

⇒ MP = 81\sqrt{81} = 9.

(i) sin θ = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= MPOP=915=35\dfrac{MP}{OP} = \dfrac{9}{15} = \dfrac{3}{5}.

Hence, sin θ = 35\dfrac{3}{5}.

(ii) cos θ = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= OMOP=1215=45\dfrac{OM}{OP} = \dfrac{12}{15} = \dfrac{4}{5}.

Hence, cos θ = 45\dfrac{4}{5}.

(iii) tan θ = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= MPOM=912=34\dfrac{MP}{OM} = \dfrac{9}{12} = \dfrac{3}{4}.

Hence, tan θ = 34\dfrac{3}{4}.

(iv) cot θ = BasePerpendicular\dfrac{\text{Base}}{\text{Perpendicular}}

= OMMP=129=43\dfrac{OM}{MP} = \dfrac{12}{9} = \dfrac{4}{3}.

Hence, cot θ = 43\dfrac{4}{3}.

(v) sec θ = HypotenuseBase\dfrac{\text{Hypotenuse}}{\text{Base}}

= OPOM=1512=54\dfrac{OP}{OM} = \dfrac{15}{12} = \dfrac{5}{4}.

Hence, sec θ = 54\dfrac{5}{4}.

(vi) cosec θ = HypotenusePerpendicular\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}

= OPMP=159=53\dfrac{OP}{MP} = \dfrac{15}{9} = \dfrac{5}{3}.

Hence, cosec θ = 53\dfrac{5}{3}.

Question 1(b)

From the figure (2) given below, find the values of :

(i) sin A

(ii) cos A

(iii) sin2 A + cos2 A

(iv) sec2 A - tan2 A

From the figure, find the values of : (i) sin A (ii) cos A (iii) sin^2 A + cos^2 A (iv) sec^2 A - tan^2 A. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In right-angled triangle ABC,

By Pythagoras theorem, we get

⇒ AB2 = AC2 + BC2

⇒ AB2 = (12)2 + (5)2

⇒ AB2 = 144 + 25

⇒ AB2 = 169

⇒ AB = 169\sqrt{169} = 13

(i) sin A = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= BCAB=513\dfrac{BC}{AB} = \dfrac{5}{13}.

Hence, sin A = 513\dfrac{5}{13}.

(ii) cos A = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= ACAB=1213\dfrac{AC}{AB} = \dfrac{12}{13}.

Hence, cos A = 1213\dfrac{12}{13}.

(iii) Substituting values of sin A and cos A in sin2 A + cos2 A :

⇒ sin2 A + cos2 A = (513)2+(1213)2\Big(\dfrac{5}{13}\Big)^2 + \Big(\dfrac{12}{13}\Big)^2

= 25169+144169\dfrac{25}{169} + \dfrac{144}{169}

= 169169\dfrac{169}{169} = 1.

Hence, sin2 A + cos2 A = 1.

(iv) sec2 A - tan2 A = (ABAC)2−(BCAC)2\Big(\dfrac{AB}{AC}\Big)^2 - \Big(\dfrac{BC}{AC}\Big)^2

= (1312)2−(512)2\Big(\dfrac{13}{12}\Big)^2 - \Big(\dfrac{5}{12}\Big)^2

= 169144−25144\dfrac{169}{144} - \dfrac{25}{144}

= 144144\dfrac{144}{144} = 1.

Hence, sec2 A - tan2 A = 1.

Question 2(a)

From the figure (1) given below, find the values of:

(i) sin B

(ii) cos C

(iii) sin B + sin C

(iv) sin B cos C + sin C cos B.

From the figure, find the values of (i) sin B (ii) cos C (iii) sin B + sin C (iv) sin B cos C + sin C cos B. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In right-angled triangle ABC,

By Pythagoras theorem, we get

⇒ BC2 = AC2 + AB2

⇒ AC2 = BC2 - AB2

⇒ AC2 = (10)2 - (6)2

⇒ AC2 = 100 - 36

⇒ AC2 = 64

⇒ AC = 64\sqrt{64}

⇒ AC = 8.

(i) sin B = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= ACBC=810=45\dfrac{AC}{BC} = \dfrac{8}{10} = \dfrac{4}{5}.

Hence, sin B = 45\dfrac{4}{5}.

(ii) cos C = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= ACBC=810=45.\dfrac{AC}{BC} = \dfrac{8}{10} = \dfrac{4}{5}.

Hence, cos C = 45\dfrac{4}{5}.

(iii) sin C = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= ABBC=610=35\dfrac{AB}{BC} = \dfrac{6}{10} = \dfrac{3}{5}.

Substituting values of sin B and sin C in sin B + sin C we get :

⇒ sin B + sin C = 45+35=75\dfrac{4}{5} + \dfrac{3}{5} = \dfrac{7}{5}.

Hence, sin B + sin C = 75\dfrac{7}{5}.

(iv) sin B = 45\dfrac{4}{5}, sin C = 35\dfrac{3}{5}, cos C = 45\dfrac{4}{5}.

cos B = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= ABBC=610=35\dfrac{AB}{BC} = \dfrac{6}{10} = \dfrac{3}{5}.

Substituting values in equation sin B cos C + sin C cos B we get :

=45×45+35×35=1625+925=2525=1.= \dfrac{4}{5} \times \dfrac{4}{5} + \dfrac{3}{5} \times \dfrac{3}{5} \\[1em] = \dfrac{16}{25} + \dfrac{9}{25} \\[1em] = \dfrac{25}{25} = 1.

Hence, sin B cos C + sin C cos B = 1.

Question 2(b)

From the figure (2) given below, find the values of :

(i) tan x

(ii) cos y

(iii) cosec2 y - cot2 y

(iv) 5sin x+3sin y−3 cot y\dfrac{5}{\text{sin x}} + \dfrac{3}{\text{sin y}} - 3\text{ cot y}.

From the figure, find the values of (i) tan x (ii) cos y (iii) cosec^2 y - cot^2 y. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

From Figure,

BD = BC - CD = 21 - 5 = 16.

In right-angled ∆ACD,

By pythagoras theorem we get,

⇒ AC2 = AD2 + CD2

⇒ AD2 = AC2 - CD2

⇒ AD2 = (13)2 - (5)2

⇒ AD2 = 169 - 25

⇒ AD2 = 144

⇒ AD = 144\sqrt{144}

⇒ AD = 12.

In right-angled ∆ABD,

By pythagoras theorem we get,

⇒ AB2 = AD2 + BD2

⇒ AB2 = 122 + 162

⇒ AB2 = 144 + 256

⇒ AB2 = 400

⇒ AB = 400\sqrt{400}

⇒ AB = 20.

(i) In right-angled ∆ACD,

tan x = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= CDAD=512\dfrac{CD}{AD} = \dfrac{5}{12}.

Hence, tan x = 512\dfrac{5}{12}.

(ii) In right-angled ∆ABD,

cos y = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= BDAB=1620=45\dfrac{BD}{AB} = \dfrac{16}{20} = \dfrac{4}{5}.

Hence, cos y = 45\dfrac{4}{5}.

(iii) In right-angled ∆ABD,

cosec y = HypotenusePerpendicular\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}

= ABAD=2012=53\dfrac{AB}{AD} = \dfrac{20}{12} = \dfrac{5}{3}.

cot y = BasePerpendicular\dfrac{\text{Base}}{\text{Perpendicular}}

= BDAD=1612=43\dfrac{BD}{AD} = \dfrac{16}{12} = \dfrac{4}{3}.

Substituting values in cosec2 y - cot2 y

=(53)2−(43)2=259−169=99=1.= \Big(\dfrac{5}{3}\Big)^2 - \Big(\dfrac{4}{3}\Big)^2 \\[1em] = \dfrac{25}{9} - \dfrac{16}{9} \\[1em] = \dfrac{9}{9} = 1.

Hence, cosec2 y - cot2 y = 1.

(iv) In right-angled ∆ACD,

sin x = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= CDAC=513\dfrac{CD}{AC} = \dfrac{5}{13}.

In right-angled ∆ABD,

sin y = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= ADAB=1220=35\dfrac{AD}{AB} = \dfrac{12}{20} = \dfrac{3}{5}.

Substituting values in 5sin x+3sin y−3 cot y\dfrac{5}{\text{sin x}} + \dfrac{3}{\text{sin y}} - 3\text{ cot y}, we get :

=5513+335−3×43=13+5−4=14.= \dfrac{5}{\dfrac{5}{13}} + \dfrac{3}{\dfrac{3}{5}} - 3 \times \dfrac{4}{3} \\[1em] = 13 + 5 - 4 \\[1em] = 14.

Hence, 5sin x+3sin y−3 cot y=14\dfrac{5}{\text{sin x}} + \dfrac{3}{\text{sin y}} - 3\text{ cot y} = 14

Question 3(a)

From the figure (1) given below, find the value of sec θ.

From the figure, find the value of sec θ. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

From figure,

⇒ BD = BC - CD = 21 - 5 = 16.

In △ADC,

⇒ AC2 = AD2 + DC2 [By pythagoras theorem]

⇒ AD2 = AC2 - DC2

⇒ AD2 = 132 - 52

⇒ AD2 = 169 - 25

⇒ AD2 = 144

⇒ AD = 144\sqrt{144}

⇒ AD = 12.

In △ABD,

⇒ AB2 = AD2 + BD2 [By pythagoras theorem]

⇒ AB2 = 122 + 162

⇒ AB2 = 144 + 256

⇒ AB2 = 400

⇒ AB = 400\sqrt{400}

⇒ AB = 20.

By formula,

sec θ = HypotenuseBase\dfrac{\text{Hypotenuse}}{\text{Base}}

= ABBD=2016=54\dfrac{AB}{BD} = \dfrac{20}{16} = \dfrac{5}{4}.

Hence, sec θ = 54\dfrac{5}{4}.

Question 3(b)

From the figure (2) given below, find the value of :

(i) sin x

(ii) cot x

(iii) cot2 x - cosec2 x

(iv) sec y

(v) tan2 y - 1cos2y\dfrac{1}{\text{cos}^2 y}

From the figure, find the value of (i) sin x (ii) cot x (iii) cot^2 x - cosec^2 x (iv) sec y. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In right angled ∆ABD,

From the figure, find the value of (i) sin x (ii) cot x (iii) cot^2 x - cosec^2 x (iv) sec y. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By pythagoras theorem we get,

⇒ AD2 = AB2 + BD2

⇒ AD2 = 32 + 42

⇒ AD2 = 9 + 16

⇒ AD2 = 25

⇒ AD = 25\sqrt{25}

⇒ AD = 5

In right angled triangle DBC,

By pythagoras theorem we get,

⇒ DC2 = DB2 + BC2

⇒ 122 = 42 + BC2

⇒ BC2 = 122 - 42

⇒ BC2 = 144 - 16

⇒ BC2 = 128

⇒ BC = 128\sqrt{128}

⇒ BC = 828\sqrt{2}.

(i) sin x = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= BDAD=45\dfrac{BD}{AD} = \dfrac{4}{5}.

Hence, sin x = 45\dfrac{4}{5}.

(ii) cot x = BasePerpendicular\dfrac{\text{Base}}{\text{Perpendicular}}

= ABBD=34\dfrac{AB}{BD} = \dfrac{3}{4}.

Hence, cot x = 34\dfrac{3}{4}.

(iii) cosec x = HypotenusePerpendicular\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}

= ADBD=54\dfrac{AD}{BD} = \dfrac{5}{4}.

Substituting values in cot2 x - cosec2 x we get :

=(34)2−(54)2=916−2516=−1616=−1.= \Big(\dfrac{3}{4}\Big)^2 - \Big(\dfrac{5}{4}\Big)^2 \\[1em] = \dfrac{9}{16} - \dfrac{25}{16} \\[1em] = -\dfrac{16}{16} \\[1em] = -1.

Hence, cot2 x - cosec2 x = -1.

(iv) sec y = HypotenuseBase\dfrac{\text{Hypotenuse}}{\text{Base}}

= CDBC=1282=322\dfrac{CD}{BC} = \dfrac{12}{8\sqrt{2}} = \dfrac{3}{2\sqrt{2}}.

Hence, sec y = 322\dfrac{3}{2\sqrt{2}}.

(v) tan y = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= BDBC=482=122\dfrac{BD}{BC} = \dfrac{4}{8\sqrt{2}} = \dfrac{1}{2\sqrt{2}}.

Given,

⇒tan2y−1cos2y⇒(122)2−sec2y⇒18−(322)2⇒18−98⇒1−98⇒−88⇒−1.\Rightarrow \text{tan}^2 y - \dfrac{1}{\text{cos}^2 y} \\[1em] \Rightarrow \Big(\dfrac{1}{2\sqrt{2}}\Big)^2 - \text{sec}^2 y \\[1em] \Rightarrow \dfrac{1}{8} - \Big(\dfrac{3}{2\sqrt{2}}\Big)^2 \\[1em] \Rightarrow \dfrac{1}{8} - \dfrac{9}{8} \\[1em] \Rightarrow \dfrac{1 - 9}{8} \\[1em] \Rightarrow \dfrac{-8}{8} \\[1em] \Rightarrow -1.

Hence, tan2y−1cos2y=−1\text{tan}^2 y - \dfrac{1}{\text{cos}^2 y} = -1

Question 4(a)

From the figure (1) given below, find the values of :

(i) 2 sin y - cos y

(ii) 2 sin x - cos x

(iii) 1 - sin x + cos y

(iv) 2 cos x - 3 sin y + 4 tan x

From the figure, find the values of (i) 2 sin y - cos y (ii) 2 sin x - cos x (iii) 1 - sin x + cos y (iv) 2 cos x - 3 sin y + 4 tan x. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In right-angled ∆BCD,

Using pythagoras theorem we get,

⇒ BC2 = BD2 + CD2

⇒ BC2 = 92 + 122

⇒ BC2 = 81 + 144

⇒ BC2 = 225

⇒ BC = 225\sqrt{225}

⇒ BC = 15.

In a right-angled ∆ABC,

Using pythagoras theorem

⇒ AC2 = AB2 + BC2

⇒ AB2 = AC2 - BC2

⇒ AB2 = 252 - 152

⇒ AB2 = 625 - 225 = 400

⇒ AB = 400\sqrt{400}

⇒ AB = 20

(i) We know that

In right-angled ∆BCD,

sin y = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= BDBC=915=35.\dfrac{BD}{BC} = \dfrac{9}{15} = \dfrac{3}{5}.

cos y = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= CDBC=1215=45.\dfrac{CD}{BC} = \dfrac{12}{15} = \dfrac{4}{5}.

Substituting values in 2 sin y - cos y we get :

⇒2 sin y - cos y=2×35−45=65−45=25.\Rightarrow \text{2 sin y - cos y} = 2 \times \dfrac{3}{5} - \dfrac{4}{5} \\[1em] = \dfrac{6}{5} - \dfrac{4}{5} \\[1em] = \dfrac{2}{5}.

Hence, 2 sin y - cos y = 25\dfrac{2}{5}.

(ii) In right-angled ∆ABC

sin x = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= BCAC=1525=35.\dfrac{BC}{AC} = \dfrac{15}{25} = \dfrac{3}{5}.

cos x = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= ABAC=2025=45.\dfrac{AB}{AC} = \dfrac{20}{25} = \dfrac{4}{5}.

Substituting the values in 2 sin x - cos x we get :

⇒2 sin x - cos x=2×35−45=65−45=25.\Rightarrow \text{2 sin x - cos x} = 2 \times \dfrac{3}{5} - \dfrac{4}{5} \\[1em] = \dfrac{6}{5} - \dfrac{4}{5} \\[1em] = \dfrac{2}{5}.

Hence, 2 sin x - cos x = 25\dfrac{2}{5}.

(iii) Substituting values we get :

⇒1 - sin x + cos y=1−35+45=5−3+45=65.\Rightarrow \text{1 - sin x + cos y} = 1 - \dfrac{3}{5} + \dfrac{4}{5} \\[1em] = \dfrac{5 - 3 + 4}{5} \\[1em] = \dfrac{6}{5}.

Hence, 1 - sin x + cos y = 65\dfrac{6}{5}.

(iv) By formula,

tan x = sin xcos x=3545=34\dfrac{\text{sin x}}{\text{cos x}} = \dfrac{\dfrac{3}{5}}{\dfrac{4}{5}} = \dfrac{3}{4}.

Substituting values we get :

⇒2 cos x - 3 sin y + 4 tan x=2×45−3×35+4×34=85−95+3=−8−9+155=145.\Rightarrow \text{2 cos x - 3 sin y + 4 tan x} = 2 \times \dfrac{4}{5} - 3 \times \dfrac{3}{5} + 4 \times \dfrac{3}{4}\\[1em] = \dfrac{8}{5} - \dfrac{9}{5} + 3 \\[1em] = -\dfrac{8 - 9 + 15}{5} \\[1em] = \dfrac{14}{5}.

Hence, 2 cos x - 3 sin y + 4 tan x = 145\dfrac{14}{5}.

Question 4(b)

In the figure (2), given below, ΔABC is right angled at B. If sin θ = 513\dfrac{5}{13} and BC = 24 cm, find AB and AC.

In the figure (2), given below, ΔABC is right angled at B. If sin θ =: Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

Given, ΔABC is a right angled at B.

sin θ = 513\dfrac{5}{13} and BC = 24 cm

⇒sin θ=PerpendicularHypotenuse⇒513=ABAC\Rightarrow \text{sin θ} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}\\[1em] \Rightarrow \dfrac{5}{13} = \dfrac{AB}{AC}

Let AB = 5k and AC = 13k.

Using pythagoras theorem,

⇒ AC2 = AB2 + BC2

⇒ (13k)2 = (5k)2 + 242

⇒ 169k2 = 25k2 + 576

⇒ 169k2 - 25k2 = 576

⇒ 144k2 = 576

⇒ k2 = 576144\dfrac{576}{144}

⇒ k2 = 4

⇒ k = 4\sqrt{4}

⇒ k = ± 2

Since, length cannot be negative.

⇒ k = 2

So, AB = 5k = 5 x 2 = 10 cm,

and AC = 13k = 13 x 2 = 26 cm.

Hence, AB = 10 cm and AC = 26 cm.

Question 5

In a right-angled triangle, it is given that angle A is an acute angle and that tan A = 512\dfrac{5}{12}. Find the values of :

(i) cos A

(ii) cosec A - cot A.

Answer

Let ABC be a right angled triangle with ∠B = 90°.

In a right-angled triangle, it is given that angle A is an acute angle and that tan A = 5/12. Find the values of : (i) cos A (ii) cosec A - cot A. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Given,

tan A = 512\dfrac{5}{12}.

By formula,

tan A = PerpendicularBase=BCAB=512\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{BC}{AB} = \dfrac{5}{12}

Let BC = 5x and AB = 12x.

From right-angled ∆ABC

By pythagoras theorem, we get

⇒ AC2 = BC2 + AB2

⇒ AC2 = (5x)2 + (12x)2

⇒ AC2 = 25x2 + 144x2

⇒ AC2 = 169x2

⇒ AC = 169x2\sqrt{169x^2}

⇒ AC = 13x.

(i) By formula,

cos A = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= ABAC=12x13x=1213\dfrac{AB}{AC} = \dfrac{12x}{13x} = \dfrac{12}{13}.

Hence, cos A = 1213\dfrac{12}{13}.

(ii) By formula,

cosec A = HypotenusePerpendicular\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}

= ACBC=13x5x=135\dfrac{AC}{BC} = \dfrac{13x}{5x} = \dfrac{13}{5}.

cot A = BasePerpendicular\dfrac{\text{Base}}{\text{Perpendicular}}

= ABBC\dfrac{AB}{BC} = 12x5x\dfrac{12x}{5x} = 125\dfrac{12}{5}.

Substituting values in cosec A - cot A, we get :

⇒cosec A - cot A=135−125=15.\Rightarrow \text{cosec A - cot A} = \dfrac{13}{5} - \dfrac{12}{5} \\[1em] = \dfrac{1}{5}.

Hence, cosec A - cot A = 15\dfrac{1}{5}.

Question 6(a)

In ∆ABC, ∠A = 90°. If AB = 7 cm and BC - AC = 1 cm, find :

(i) sin C

(ii) tan B

Answer

(a) In right ∆ABC

In ∆ABC, ∠A = 90°. If AB = 7 cm and BC - AC = 1 cm, find (i) sin C (ii) tan B. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

∠A = 90°

AB = 7 cm

BC - AC = 1 cm

⇒ BC = 1 + AC

We know that,

⇒ BC2 = AB2 + AC2 [By pythagoras theorem]

⇒ (1 + AC)2 = AB2 + AC2

⇒ 1 + AC2 + 2AC = 72 + AC2

⇒ 1 + AC2 + 2AC = 49 + AC2

⇒ 2AC = 49 - 1

⇒ 2AC = 48

⇒ AC = 482\dfrac{48}{2} = 24 cm.

∴ BC = 1 + AC = 25 cm.

(i) sin C = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= ABBC=725\dfrac{AB}{BC} = \dfrac{7}{25}

Hence, sin C = 725\dfrac{7}{25}.

(ii) tan B = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= ACAB=247\dfrac{AC}{AB} = \dfrac{24}{7}

Hence, tan B = 247\dfrac{24}{7}.

Question 6(b)

In △PQR, ∠Q = 90°. If PQ = 40 cm and PR + QR = 50 cm, find :

(i) sin P

(ii) cos P

(iii) tan R

Answer

In right ∆PQR,

∠Q = 90°

PQ = 40 cm

In △PQR, ∠Q = 90°. If PQ = 40 cm and PR + QR = 50 cm, find (i) sin P (ii) cos P (iii) tan R. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Given,

⇒ PR + QR = 50 cm

⇒ PR = 50 - QR

Using Pythagoras theorem we get,

⇒ PR2 = PQ2 + QR2

⇒ (50 - QR)2 = (40)2 + QR2

⇒ 502 + QR2 - 100QR = 1600 + QR2

⇒ QR2 - QR2 - 100QR = 1600 - 2500

⇒ -100QR = -900

⇒ 100QR = 900

⇒ QR = 900100\dfrac{900}{100} = 9.

∴ PR = 50 - QR = 50 - 9 = 41.

(i) sin P = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= QRPR=941\dfrac{QR}{PR} = \dfrac{9}{41}.

Hence, sin P = 941\dfrac{9}{41}.

(ii) cos P = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= PQPR=4041\dfrac{PQ}{PR} = \dfrac{40}{41}.

Hence, cos P = 4041\dfrac{40}{41}.

(iii) tan R = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= PQQR=409\dfrac{PQ}{QR} = \dfrac{40}{9}.

Hence, tan R = 409\dfrac{40}{9}.

Question 7

In △ABC, AB = AC = 15 cm, BC = 18 cm. Find

(i) cos ∠ABC

(ii) sin ∠ACB.

Answer

Draw AD perpendicular to BC.

D is mid-point of BC [∵ Perpendicular drawn to the unequal side of an isosceles triangle from the apex vertex bisects the side]

∴ BD = DC = 9 cm.

In △ABC, AB = AC = 15 cm, BC = 18 cm. Find (i) cos ∠ABC (ii) sin ∠ACB. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

In right-angled triangle ABD,

⇒ AB2 = AD2 + BD2

⇒ 152 = AD2 + 92

⇒ AD2 = (15)2 - (9)2

⇒ AD2 = 225 - 81

⇒ AD2 = 144

⇒ AD = 144\sqrt{144} = 12 cm.

(i) cos ∠ABC = BDAB\dfrac{BD}{AB}

= 915=35\dfrac{9}{15} = \dfrac{3}{5}.

Hence, cos ∠ABC = 35\dfrac{3}{5}.

(ii) sin ∠ACB = ADAC\dfrac{AD}{AC}

= 1215=45\dfrac{12}{15} = \dfrac{4}{5}.

Hence, sin ∠ACB = 45\dfrac{4}{5}.

Question 8(a)

In the figure (1) given below, ∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm. Find

(i) sin C

(ii) tan B

(iii) tan C - cot B.

In the figure, ∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm. Find (i) sin C (ii) tan B (iii) tan C - cot B. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

Draw AD perpendicular to BC.

D is the mid point of BC [∵ Perpendicular drawn to the unequal side of an isosceles triangle from the apex vertex bisects the side]

So, BD = CD = 3 cm.

In the figure, ∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm. Find (i) sin C (ii) tan B (iii) tan C - cot B. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

In right-angled ∆ABD,

Using pythagoras theorem we get :

⇒ AB2 = AD2 + BD2

⇒ AD2 = AB2 - BD2

⇒ AD2 = 52 - 32

⇒ AD2 = 25 - 9

⇒ AD2 = 16

⇒ AD = 16\sqrt{16}

⇒ AD = 4 cm.

(i) In right-angled ∆ACD,

sin C = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= ADAC=45\dfrac{AD}{AC} = \dfrac{4}{5}.

Hence, sin C = 45\dfrac{4}{5}.

(ii) In right-angled ∆ABD,

tan B = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= ADBD=43\dfrac{AD}{BD} = \dfrac{4}{3}.

Hence, tan B = 43\dfrac{4}{3}.

(iii) In right-angled ∆ACD,

tan C = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= ADCD=43\dfrac{AD}{CD} = \dfrac{4}{3}.

In right-angled ∆ABD,

cot B = basePerpendicular\dfrac{\text{base}}{\text{Perpendicular}}

= BDAD=34\dfrac{BD}{AD} = \dfrac{3}{4}.

Substituting values in tan C - cot B we get :

tan C−cot B=43−34=16−912=712.\text{tan C} - \text{cot B} = \dfrac{4}{3} - \dfrac{3}{4} \\[1em] = \dfrac{16 - 9}{12} \\[1em] = \dfrac{7}{12}.

Hence, tan C - cot B = 712\dfrac{7}{12}.

Question 8(b)

In the figure (2) given below, △ABC is right-angled at B. Given that ∠ACB = θ, side AB = 2 units and side BC = 1 unit, find the value of sin2 θ + tan2 θ.

In the figure, △ABC is right-angled at B. Given that ∠ACB = θ, side AB = 2 units and side BC = 1 unit, find the value of sin^2 θ + tan^2 θ. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ AC2 = (2)2 + (1)2

⇒ AC2 = 4 + 1 = 5

⇒ AC = 5\sqrt{5}.

Calculating sin θ, we get :

sin θ=PerpendicularHypotenuse=ABAC=25.\text{sin θ} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{AB}{AC} \\[1em] = \dfrac{2}{\sqrt{5}}.

Calculating tan θ, we get :

tan θ=PerpendicularBase=ABBC=21=2.\text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AB}{BC} \\[1em] = \dfrac{2}{1} = 2.

Substituting value of sin θ and tan θ in sin2 θ + tan2 θ, we get :

sin2 θ+tan2 θ=(25)2+22=45+4=4+205=245=445.\text{sin}^2 \text{ θ} + \text{tan}^2 \text{ θ} = \Big(\dfrac{2}{\sqrt{5}}\Big)^2 + 2^2 \\[1em] = \dfrac{4}{5} + 4 \\[1em] = \dfrac{4 + 20}{5} \\[1em] = \dfrac{24}{5} = 4\dfrac{4}{5}.

Hence, sin2 θ + tan2 θ = 4454\dfrac{4}{5}.

Question 8(c)

In the figure (3) given below, AD is perpendicular to BC, BD = 15 cm, sin B = 45\dfrac{4}{5} and tan C = 1.

(i) Calculate the lengths of AD, AB, DC and AC.

(ii) Show that tan2 B−1cos2 B\text{tan}^2 \text{ B} - \dfrac{1}{\text{cos}^2 \text{ B}} = -1.

In the figure, AD is perpendicular to BC, BD = 15 cm, sin B = 4/5 and tan C = 1. (i) Calculate the lengths of AD, AB, DC and AC. (ii) Show that tan^2 B - 1/cos^2 B = -1. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

(i) From figure,

tan C = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}} = ADDC\dfrac{AD}{DC}

⇒ 1 = ADDC\dfrac{AD}{DC}

⇒ AD = DC.

sin B = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = ADAB\dfrac{AD}{AB}

⇒ 45\dfrac{4}{5} = ADAB\dfrac{AD}{AB}

Let AD = 4k and AB = 5k.

In right angle triangle ABD,

⇒ AB2 = AD2 + BD2

⇒ (5k)2 = (4k)2 + 152

⇒ 25k2 = 16k2 + 225

⇒ 9k2 = 225

⇒ k2 = 2259\dfrac{225}{9} = 25

⇒ k = 25\sqrt{25} = 5.

AD = 4k = 4 × 5 = 20

AB = 5k = 5 × 5 = 25.

DC = AD = 20.

In right angle triangle ADC,

⇒ AC2 = AD2 + DC2

⇒ AC2 = 202 + 202

⇒ AC2 = 400 +400

⇒ AC2 = 800

⇒ AC = 800=202\sqrt{800} = 20\sqrt{2}.

Hence, AD = 20, AB = 25, DC = 20 and AC = 20220\sqrt{2}.

(ii) Calculating tan B, we get :

tan B=PerpendicularBase=ADBD=2015=43.\text{tan B} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AD}{BD} \\[1em] = \dfrac{20}{15} = \dfrac{4}{3}.

Calculating cos B, we get :

cos B=BaseHypotenuse=BDAB=1525=35.\text{cos B} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{BD}{AB} \\[1em] = \dfrac{15}{25} = \dfrac{3}{5}.

Substituting value of tan B and cos B in L.H.S. of the equation, tan2 B−1cos2 B\text{tan}^2 \text{ B} - \dfrac{1}{\text{cos}^2 \text{ B}} = -1, we get :

⇒tan2B−1cos2B=(43)2−1(35)2=169−1925=169−259=−99=−1.\Rightarrow \text{tan}^2 B - \dfrac{1}{\text{cos}^2 B} = \Big(\dfrac{4}{3}\Big)^2 - \dfrac{1}{\Big(\dfrac{3}{5}\Big)^2} \\[1em] = \dfrac{16}{9} - \dfrac{1}{\dfrac{9}{25}} \\[1em] = \dfrac{16}{9} - \dfrac{25}{9} \\[1em] = \dfrac{-9}{9} \\[1em] = -1.

Hence, proved that tan2 B - 1cos2B\dfrac{1}{\text{cos}^2 B} = -1.

Question 9

If sin θ = 35\dfrac{3}{5} and θ is acute angle, find

(i) cos θ

(ii) tan θ.

Answer

(i) By formula,

⇒ sin2 θ + cos2 θ = 1

Substituting values we get,

⇒(35)2+cos2 θ=1⇒cos2 θ=1−925⇒cos2 θ=25−925⇒cos2 θ=1625⇒cos θ=1625⇒cos θ=±45.\Rightarrow \Big(\dfrac{3}{5}\Big)^2 + \text{cos}^2 \text{ θ} = 1 \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = 1 - \dfrac{9}{25} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \dfrac{25 - 9}{25} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \dfrac{16}{25} \\[1em] \Rightarrow \text{cos θ} = \sqrt{\dfrac{16}{25}} \\[1em] \Rightarrow \text{cos θ} = \pm \dfrac{4}{5}.

Since, θ is an acute angle and value of cos is positive in first quadrant.

∴ cos θ = 45\dfrac{4}{5}.

Hence, cos θ = 45\dfrac{4}{5}.

(ii) By formula,

tan θ = sin θcos θ\dfrac{\text{sin θ}}{\text{cos θ}}

= 3545=34\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}} = \dfrac{3}{4}.

Hence, tan θ = 34\dfrac{3}{4}.

Question 10

Given that tan θ = 512\dfrac{5}{12} and θ is an acute angle, find sin θ and cos θ.

Answer

By formula,

sec2 θ = 1 + tan2 θ

Substituting values we get,

⇒sec2 θ=1+(512)2⇒sec2 θ=1+25144⇒sec2 θ=144+25144⇒sec2 θ=169144⇒sec θ=169144⇒sec θ=±1312.\Rightarrow \text{sec}^2 \text{ θ} = 1 + \Big(\dfrac{5}{12}\Big)^2 \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = 1 + \dfrac{25}{144} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \dfrac{144 + 25}{144} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \dfrac{169}{144} \\[1em] \Rightarrow \text{sec θ} = \sqrt{\dfrac{169}{144}} \\[1em] \Rightarrow \text{sec θ} = \pm \dfrac{13}{12}.

Since, θ is an acute angle and value of sec is positive in first quadrant.

∴ sec θ = 1312\dfrac{13}{12}.

By formula,

cos θ = 1sec θ=11312=1213.\dfrac{1}{\text{sec θ}} = \dfrac{1}{\dfrac{13}{12}} = \dfrac{12}{13}.

sin2 θ + cos2 θ = 1

Substituting values we get,

⇒sin2 θ+(1213)2=1⇒sin2 θ=1−(1213)2⇒sin2 θ=1−144169⇒sin2 θ=169−144169⇒sin2 θ=25169⇒sin θ=25169⇒sin θ=±513.\Rightarrow \text{sin}^2 \text{ θ} + \Big(\dfrac{12}{13}\Big)^2 = 1 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = 1 - \Big(\dfrac{12}{13}\Big)^2\\[1em] \Rightarrow \text{sin}^2 \text{ θ} = 1 - \dfrac{144}{169} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \dfrac{169 - 144}{169} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \dfrac{25}{169} \\[1em] \Rightarrow \text{sin θ} = \sqrt{\dfrac{25}{169}} \\[1em] \Rightarrow \text{sin θ} = \pm \dfrac{5}{13}.

Since, θ is an acute angle and value of sin is positive in first quadrant.

∴ sin θ = 513\dfrac{5}{13}.

Hence, sin θ = 513\dfrac{5}{13} and cos θ = 1213\dfrac{12}{13}.

Question 11

If sin θ = 610\dfrac{6}{10}, find the value of cos θ + tan θ.

Answer

Given,

⇒sin θ=610⇒sin2 θ=36100⇒1−cos2 θ=36100⇒cos2 θ=1−36100⇒cos2 θ=100−36100⇒cos2 θ=64100⇒cos θ=64100⇒cos θ=810tan θ=sin θcos θ=610810=68.\Rightarrow \text{sin θ} = \dfrac{6}{10} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \dfrac{36}{100} \\[1em] \Rightarrow 1 - \text{cos}^2 \text{ θ} = \dfrac{36}{100} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = 1 - \dfrac{36}{100} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \dfrac{100 - 36}{100} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \dfrac{64}{100} \\[1em] \Rightarrow \text{cos θ} = \sqrt{\dfrac{64}{100}} \\[1em] \Rightarrow \text{cos θ} = \dfrac{8}{10} \\[1.5em] \text{tan θ} = \dfrac{\text{sin θ}}{\text{cos θ}} \\[1em] = \dfrac{\dfrac{6}{10}}{\dfrac{8}{10}} = \dfrac{6}{8}.

Substituting values in cos θ + tan θ we get,

cos θ + tan θ =810+68=32+3040=6240=3120=11120\text{cos θ + tan θ }= \dfrac{8}{10} + \dfrac{6}{8} \\[1em] = \dfrac{32 + 30}{40} \\[1em] = \dfrac{62}{40} \\[1em] = \dfrac{31}{20} \\[1em] = 1\dfrac{11}{20}

Hence, cos θ + tan θ = 11120.1\dfrac{11}{20}.

Question 12

If tan θ = 43\dfrac{4}{3}, find the value of sin θ + cos θ (both sin θ and cos θ are positive).

Answer

Given,

⇒tan θ=43⇒tan2 θ=169⇒sec2 θ=1+tan2θ⇒sec2 θ=1+169⇒sec2 θ=9+169⇒sec2 θ=259⇒1cos2 θ=259⇒cos2 θ=925⇒cos θ=925⇒cos θ=±35⇒cos θ=35[Given, cos θ is positive.].\phantom{\Rightarrow} \text{tan θ} = \dfrac{4}{3} \\[1em] \Rightarrow \text{tan}^2 \text{ θ} = \dfrac{16}{9} \\[1em] \phantom{\Rightarrow} \text{sec}^2 \text{ θ} = 1 + \text{tan}^2 θ \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = 1 + \dfrac{16}{9} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \dfrac{9 + 16}{9} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \dfrac{25}{9} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 \text{ θ}} = \dfrac{25}{9} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \dfrac{9}{25} \\[1em] \Rightarrow \text{cos θ} = \sqrt{\dfrac{9}{25}} \\[1em] \Rightarrow \text{cos θ} = \pm\dfrac{3}{5} \\[1em] \Rightarrow \text{cos θ} = \dfrac{3}{5} [\text{Given, cos θ is positive.}].

By formula,

sin2 θ + cos2 θ = 1

Substituting values in sin2 θ + cos2 θ = 1 we get :

⇒sin2 θ+(35)2=1⇒sin2 θ=1−925⇒sin2 θ=25−925⇒sin2 θ=1625⇒sin θ=1625⇒sin θ=±45⇒sin θ=45 [Given, sin θ is positive.].\Rightarrow \text{sin}^2 \text{ θ} + \Big(\dfrac{3}{5}\Big)^2 = 1 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = 1 - \dfrac{9}{25} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \dfrac{25 - 9}{25} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \dfrac{16}{25} \\[1em] \Rightarrow \text{sin θ} = \sqrt{\dfrac{16}{25}} \\[1em] \Rightarrow \text{sin θ} = \pm\dfrac{4}{5} \\[1em] \Rightarrow \text{sin θ} = \dfrac{4}{5} \space [\text{Given, sin θ is positive.}].

Substituting values in sin θ + cos θ we get :

sin θ + cos θ = 45+35=75=125.\dfrac{4}{5} + \dfrac{3}{5} = \dfrac{7}{5} = 1\dfrac{2}{5}.

Hence, sin θ + cos θ = 1251\dfrac{2}{5}.

Question 13

If cosec θ = 5\sqrt{5} and θ is less than 90°, find the value of cot θ - cos θ.

Answer

Let ABC be a right angle triangle with ∠C = θ and ∠B = 90°.

If cosec θ = √5 and θ is less than 90°, find the value of cot θ - cos θ. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By formula,

cosec θ = HypotenusePerpendicular=ACAB\dfrac{\text{Hypotenuse}}{\text{Perpendicular}} = \dfrac{AC}{AB}.

Substituting values we get :

⇒51=ACAB\Rightarrow \dfrac{\sqrt{5}}{1} = \dfrac{AC}{AB}

Let AC = 5\sqrt{5}k and AB = k.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ (5\sqrt{5}k)2 = BC2 + k2

⇒ 5k2 = BC2 + k2

⇒ BC2 = 5k2 - k2

⇒ BC2 = 4k2

⇒ BC = 4k2\sqrt{4\text{k}^2}

⇒ BC = 2k.

cot θ = BasePerpendicular\dfrac{\text{Base}}{\text{Perpendicular}}

= BCAB\dfrac{BC}{AB} = 2kk\dfrac{2k}{k} = 21\dfrac{2}{1}

cos θ = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= BCAC\dfrac{BC}{AC} = 2k5k\dfrac{2k}{\sqrt{5}k} = 25\dfrac{2}{\sqrt{5}}

∴cot θ - cos θ=21−25=25−25=2(5−1)5.\therefore \text{cot θ - cos θ} = \dfrac{2}{1} - \dfrac{2}{\sqrt{5}} \\[1em] = \dfrac{2\sqrt{5} - 2}{\sqrt{5}} \\[1em] = \dfrac{2(\sqrt{5} - 1)}{\sqrt{5}}.

Hence, cot θ - cos θ = 2(5−1)5.\dfrac{2(\sqrt{5} - 1)}{\sqrt{5}}.

Question 14

Given sin θ = pq\dfrac{p}{q}, find cos θ + sin θ in terms of p and q.

Answer

Let ABC be a right angle triangle with ∠C = θ and ∠B = 90°.

Given sin θ = p/q, find cos θ + sin θ in terms of p and q. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By formula,

sin θ = PerpendicularHypotenuse=ABAC\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{AB}{AC} ........(1)

Given,

sin θ = pq\dfrac{p}{q} .........(2)

From (1) and (2) we get,

ABAC=pq\dfrac{AB}{AC} = \dfrac{p}{q}

Let AB = px and AC = qx.

In right angled triangle ABC,

By pythagoras theorem,

⇒ AC2 = AB2 + BC2

⇒ BC2 = AC2 - AB2

⇒ BC2 = (qx)2 - (px)2

⇒ BC2 = q2x2 - p2x2

⇒ BC2 = x2(q2 - p2)

⇒ BC = x2(q2−p2)\sqrt{x^2(q^2 - p^2)}

⇒ BC = x(q2−p2)x\sqrt{(q^2 - p^2)}

In right angled triangle ABC,

By formula,

cos θ = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= BCAC=xq2−p2qx=q2−p2q\dfrac{BC}{AC} = \dfrac{x\sqrt{q^2 - p^2}}{qx} = \dfrac{\sqrt{q^2 - p^2}}{q}.

Substituting values in cos θ + sin θ we get,

⇒cos θ + sin θ=q2−p2q+pq=p+q2−p2q.\Rightarrow \text{cos θ + sin θ} = \dfrac{\sqrt{q^2 - p^2}}{q} + \dfrac{p}{q} \\[1em] = \dfrac{p + \sqrt{q^2 - p^2}}{q}.

Hence, cos θ + sin θ = p+q2−p2q.\dfrac{p + \sqrt{q^2 - p^2}}{q}.

Question 15

If θ is an acute angle and tan θ = 815\dfrac{8}{15}, find the value of sec θ + cosec θ.

Answer

Given,

tan θ = 815\dfrac{8}{15} ..........(1)

Let ABC be a right angle triangle with ∠C = θ and ∠B = 90°.

If θ is an acute angle and tan θ = 8/15, find the value of sec θ + cosec θ. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By formula,

tan θ = PerpendicularBase=ABBC\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{BC} .........(2)

From (1) and (2) we get,

ABBC=815\dfrac{AB}{BC} = \dfrac{8}{15}

Let AB = 8x and BC = 15x.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ AC2 = (8x)2 + (15x)2

⇒ AC2 = 64x2 + 225x2

⇒ AC2 = 289x2

⇒ AC = 289x2\sqrt{289x^2}

⇒ AC = 17x.

By formula,

sec θ = HypotenuseBase\dfrac{\text{Hypotenuse}}{\text{Base}}

= ACBC=17x15x=1715\dfrac{AC}{BC} = \dfrac{17x}{15x} = \dfrac{17}{15}.

cosec θ = HypotenusePerpendicular\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}

= ACAB=17x8x=178\dfrac{AC}{AB} = \dfrac{17x}{8x} = \dfrac{17}{8}.

Substituting value in sec θ + cosec θ we get :

⇒sec θ + cosec θ=1715+178=136+255120=391120=331120.\Rightarrow \text{sec θ + cosec θ} = \dfrac{17}{15} + \dfrac{17}{8} \\[1em] = \dfrac{136 + 255}{120} \\[1em] = \dfrac{391}{120} \\[1em] = 3\dfrac{31}{120}.

Hence, sec θ + cosec θ = 3311203\dfrac{31}{120}.

Question 16

Given A is an acute angle and 13 sin A = 5, evaluate :

5 sin A - 2 cos Atan A\dfrac{\text{5 sin A - 2 cos A}}{\text{tan A}}

Answer

Let triangle ABC be a right-angled triangle at B and A is an acute angle.

Given A is an acute angle and 13 sin A = 5, evaluate (5 sin A - 2 cos A)/tan A. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Given,

⇒ 13 sin A = 5

⇒ sin A = 513\dfrac{5}{13}

⇒ BCAC=513\dfrac{BC}{AC} = \dfrac{5}{13}

Let BC = 5k and AC = 13k.

In right angled triangle ABC,

Using pythagoras theorem,

⇒ AC2 = AB2 + BC2

⇒ AB2 = AC2 - BC2

⇒ AB2 = (13k)2 - (5k)2

⇒ AB2 = 169k2 - 25k2

⇒ AB2 = 144k2

⇒ AB = 144k2\sqrt{144k^2}

⇒ AB = 12k.

By formula,

cos A = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= ABAC=12k13k=1213\dfrac{AB}{AC} = \dfrac{12k}{13k} = \dfrac{12}{13}.

tan A = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= BCAB=5k12k=512\dfrac{BC}{AB} = \dfrac{5k}{12k} = \dfrac{5}{12}.

Substituting values we get,

=5 sin A - 2 cos Atan A=5×513−2×1213512=2513−2413512=113512=1265.\phantom{=} \dfrac{\text{5 sin A - 2 cos A}}{\text{tan A}} \\[1em] = \dfrac{5 \times \dfrac{5}{13} - 2 \times \dfrac{12}{13}}{\dfrac{5}{12}} \\[1em] = \dfrac{\dfrac{25}{13} - \dfrac{24}{13}}{\dfrac{5}{12}} \\[1em] = \dfrac{\dfrac{1}{13}}{\dfrac{5}{12}} \\[1em] = \dfrac{12}{65}.

Hence, 5 sin A - 2 cos Atan A=1265\dfrac{\text{5 sin A - 2 cos A}}{\text{tan A}} = \dfrac{12}{65}.

Question 17

Given A is an acute angle and cosec A = 2\sqrt{2}, find the value of

2 sin2A+3 cot2Atan2A−cos2A\dfrac{\text{2 sin}^2 A + 3\text{ cot}^2 A}{\text{tan}^2 A - \text{cos}^2 A}

Answer

Let triangle ABC be a right-angled at B and A is a acute angle.

Given A is an acute angle and cosec A = √2, find the value of (2 sin^2 A + 2 cot^2 A)/(tan^2 A - cos^2 A). Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Given,

cosec A = 2\sqrt{2}

⇒ cosec A = HypotenusePerpendicular\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}

= ACBC=21\dfrac{AC}{BC} = \dfrac{\sqrt{2}}{1}

Let AC = 2\sqrt{2}k and BC = k.

In right angled triangle ABC,

By using pythagoras theorem,

AC2 = AB2 + BC2

(2k\sqrt{2}k)2 = AB2 + k2

2k2 = AB2 + k2

AB2 = 2k2 - k2

AB2 = k2

AB = k\sqrt{\text{k}} = k.

By formula,

sin A = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= BCAC=k2k=12\dfrac{BC}{AC} = \dfrac{k}{\sqrt{2}k} = \dfrac{1}{\sqrt{2}}.

tan A = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= BCAB=kk=1\dfrac{BC}{AB} = \dfrac{k}{k} = 1.

cot A = 1tan A=11=1\dfrac{1}{\text{tan A}} = \dfrac{1}{1} = 1.

cos A = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= ABAC=k2k=12\dfrac{AB}{AC} = \dfrac{k}{\sqrt{2}k} = \dfrac{1}{\sqrt{2}}.

Substituting these values in 2 sin2A+3 cot2Atan2A−cos2A\dfrac{\text{2 sin}^2 A + 3\text{ cot}^2 A}{\text{tan}^2 A - \text{cos}^2 A} we get,

⇒2×(12)2+3×1212−(12)2⇒2×12+3×11−12⇒1+312⇒4×2⇒8.\Rightarrow \dfrac{2 \times \Big(\dfrac{1}{\sqrt{2}}\Big)^2 + 3 \times 1^2}{1^2 - \Big(\dfrac{1}{\sqrt{2}}\Big)^2} \\[1em] \Rightarrow \dfrac{2 \times \dfrac{1}{2} + 3 \times 1}{1 - \dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{1 + 3}{\dfrac{1}{2}} \\[1em] \Rightarrow 4 \times 2 \\[1em] \Rightarrow 8.

Hence, 2 sin2A+3 cot2Atan2A−cos2A\dfrac{\text{2 sin}^2 A + 3\text{ cot}^2 A}{\text{tan}^2 A - \text{cos}^2 A} = 8.

Question 18

The diagonals AC and BD of a rhombus ABCD meet at O. If AC = 8 cm and BD = 6 cm, find sin ∠OCD.

Answer

Since, diagonals of rhombus bisect each other.

∴ O is the mid point of AC.

The diagonals AC and BD of a rhombus ABCD meet at O. If AC = 8 cm and BD = 6 cm, find sin ∠OCD. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

In right angled ∆COD,

⇒ CD2 = OC2 + OD2 [By pythagoras theorem]

⇒ CD2 = 42 + 32

⇒ CD2 = 16 + 9

⇒ CD2 = 25

⇒ CD = 25\sqrt{25}

⇒ CD = 5 cm.

sin ∠OCD = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

sin ∠OCD = ODCD=35\dfrac{OD}{CD} = \dfrac{3}{5}.

Hence, sin ∠OCD = 35\dfrac{3}{5}.

Question 19

If tan θ = 512\dfrac{5}{12}, find the value of (cos θ + sin θ)(cos θ - sin θ)\dfrac{\text{(cos θ + sin θ)}}{\text{(cos θ - sin θ)}}.

Answer

Solving,

⇒(cos θ + sin θ)(cos θ - sin θ)[Dividing numerator and denominator by cos θ]⇒(cos θ + sin θ)cos θ(cos θ - sin θ)cos θ⇒cos θcos θ+sin θcos θcos θcos θ−sin θcos θ⇒1+tan θ1−tan θ[As tan θ=sin θcos θ]Substituting values we get⇒1+5121−512⇒12+51212−512⇒1712712⇒177=237.\Rightarrow \dfrac{\text{(cos θ + sin θ)}}{\text{(cos θ - sin θ)}} \\[1em] [\text{Dividing numerator and denominator by cos θ}] \\[1em] \Rightarrow \dfrac{\dfrac{\text{(cos θ + sin θ)}}{\text{cos θ}}}{\dfrac{\text{(cos θ - sin θ)}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{cos θ}}{\text{cos θ}} + \dfrac{\text{sin θ}}{\text{cos θ}}}{\dfrac{\text{cos θ}}{\text{cos θ}} - \dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{1 + \text{tan θ}}{1 - \text{tan θ}} [\text{As tan θ} = \dfrac{\text{sin θ}}{\text{cos θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \dfrac{1 + \dfrac{5}{12}}{1 - \dfrac{5}{12}} \\[1em] \Rightarrow \dfrac{\dfrac{12 + 5}{12}}{\dfrac{12 - 5}{12}} \\[1em] \Rightarrow \dfrac{\dfrac{17}{12}}{\dfrac{7}{12}} \\[1em] \Rightarrow \dfrac{17}{7} = 2\dfrac{3}{7}.

Hence, (cos θ + sin θ)(cos θ - sin θ)=237\dfrac{\text{(cos θ + sin θ)}}{\text{(cos θ - sin θ)}} = 2\dfrac{3}{7}.

Question 20

Given 5 cos A - 12 sin A = 0, find the value of sin A + cos A2 cos A - sin A\dfrac{\text{sin A + cos A}}{\text{2 cos A - sin A}}

Answer

Given,

⇒ 5 cos A - 12 sin A = 0

⇒ 5 cos A = 12 sin A

⇒ sin Acos A=512\dfrac{\text{sin A}}{\text{cos A}} = \dfrac{5}{12}

⇒ tan A = 512\dfrac{5}{12}.

We need to find the value of

(sin A + cos A)(2cos A - sin A)[Dividing numerator and denominator by cos θ]⇒(sin A + cos A)cos A(2cos A - sin A)cos A⇒sin Acos A+cos Acos A2cos Acos A−sin Acos A⇒tan A + 12−tan A [∵tan A=sin Acos A]Substituting values we get⇒512+12−512⇒5+121224−512⇒17121912⇒17×1212×19⇒1719.\dfrac{\text{(sin A + cos A)}}{\text{(2cos A - sin A)}} \\[1em] [\text{Dividing numerator and denominator by cos θ}] \\[1em] \Rightarrow \dfrac{\dfrac{\text{(sin A + cos A)}}{\text{cos A}}}{\dfrac{\text{(2cos A - sin A)}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{cos A}}}{\dfrac{\text{2cos A}}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\text{tan A + 1}}{{2 - \text{tan A}}} \space [\because \text{tan A} = \dfrac{\text{sin A}}{\text{cos A}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \dfrac{\dfrac{5}{12} + 1}{2 - \dfrac{5}{12}} \\[1em] \Rightarrow \dfrac{\dfrac{5 + 12}{12}}{\dfrac{24 - 5}{12}} \\[1em] \Rightarrow \dfrac{\dfrac{17}{12}}{\dfrac{19}{12}} \\[1em] \Rightarrow \dfrac{17 \times 12}{12 \times 19} \\[1em] \Rightarrow \dfrac{17}{19}.

Hence, sin A + cos A2 cos A - sin A=1719\dfrac{\text{sin A + cos A}}{\text{2 cos A - sin A}} = \dfrac{17}{19}.

Question 21

If tan θ = pq\dfrac{p}{q}, find the value of (p sin θ - q cos θ)(p sin θ + q cos θ)\dfrac{(\text{p sin θ - q cos θ})}{(\text{p sin θ + q cos θ)}}.

Answer

Given,

(p sin θ - q cos θ)(p sin θ + q cos θ)[Dividing numerator and denominator by cos θ]⇒(p sin θ - q cos θ)cos θ(p sin θ + q cos θ)cos θ⇒p sin θcos θ−q cos θcos θp sin θcos θ+q cos θcos θ⇒p tan θ - qp tan θ + q [∵tan θ=sin θcos θ]Substituting values we get⇒p×pq−qp×pq+q⇒p2q−qp2q+q⇒p2−q2qp2+q2q⇒p2−q2p2+q2.\dfrac{(\text{p sin θ - q cos θ})}{(\text{p sin θ + q cos θ)}} \\[1em] [\text{Dividing numerator and denominator by cos θ}] \\[1em] \Rightarrow \dfrac{\dfrac{(\text{p sin θ - q cos θ})}{\text{cos θ}}}{\dfrac{(\text{p sin θ + q cos θ)}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{p sin θ}}{\text{cos θ}} - \dfrac{\text{q cos θ}}{\text{cos θ}}}{\dfrac{\text{p sin θ}}{\text{cos θ}} + \dfrac{\text{q cos θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\text{p tan θ - q}}{{\text{p tan θ + q}}} \space [\because \text{tan θ} = \dfrac{\text{sin θ}}{\text{cos θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \dfrac{p \times \dfrac{p}{q} - q}{p \times \dfrac{p}{q} + q} \\[1em] \Rightarrow \dfrac{\dfrac{p^2}{q} - q}{\dfrac{p^2}{q} + q} \\[1em] \Rightarrow \dfrac{\dfrac{p^2 - q^2}{q}}{\dfrac{p^2 + q^2}{q}} \\[1em] \Rightarrow \dfrac{p^2 - q^2}{p^2 + q^2}.

Hence, (p sin θ - q cos θ)(p sin θ + q cos θ)=p2−q2p2+q2.\dfrac{(\text{p sin θ - q cos θ})}{(\text{p sin θ + q cos θ)}} = \dfrac{p^2 - q^2}{p^2 + q^2}.

Question 22

If 3 cot θ = 4, find the value of 5 sin θ - 3 cos θ5 sin θ + 3 cos θ\dfrac{\text{5 sin θ - 3 cos θ}}{\text{5 sin θ + 3 cos θ}}.

Answer

Given,

⇒ 3 cot θ = 4

⇒ cot θ = 43\dfrac{4}{3}

5 sin θ - 3 cos θ5 sin θ + 3 cos θ[Dividing numerator and denominator by sin θ]⇒(5 sin θ - 3 cos θ)sin θ(5 sin θ + 3 cos θ)sin θ⇒5 sin θsin θ−3 cos θsin θ5 sin θsin θ+3 cos θsin θ⇒5 - 3 cot θ5+3 cot θ [∵cot θ=cos θsin θ]Substituting values we get⇒5−3×435+3×43⇒5−45+4⇒19.\dfrac{\text{5 sin θ - 3 cos θ}}{\text{5 sin θ + 3 cos θ}} \\[1em] [\text{Dividing numerator and denominator by sin θ}] \\[1em] \Rightarrow \dfrac{\dfrac{\text{(5 sin θ - 3 cos θ)}}{\text{sin θ}}}{\dfrac{\text{(5 sin θ + 3 cos θ)}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{5 sin θ}}{\text{sin θ}} - \dfrac{\text{3 cos θ}}{\text{sin θ}}}{\dfrac{\text{5 sin θ}}{\text{sin θ}} + \dfrac{\text{3 cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{\text{5 - 3 cot θ}}{{5 + \text{3 cot θ}}} \space [\because \text{cot θ} = \dfrac{\text{cos θ}}{\text{sin θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \dfrac{5 - 3 \times \dfrac{4}{3}}{5 + 3\times \dfrac{4}{3}} \\[1em] \Rightarrow \dfrac{5 - 4}{5 + 4} \\[1em] \Rightarrow \dfrac{1}{9}.

Hence, 5 sin θ - 3 cos θ5 sin θ + 3 cos θ=19\dfrac{\text{5 sin θ - 3 cos θ}}{\text{5 sin θ + 3 cos θ}} = \dfrac{1}{9}

Question 23(i)

If 5 cos θ - 12 sin θ = 0, find the value of sin θ + cos θ2 cos θ - sin θ\dfrac{\text{sin θ + cos θ}}{\text{2 cos θ - sin θ}}.

Answer

(i) Given,

⇒ 5 cos θ - 12 sin θ = 0

⇒ 5 cos θ = 12 sin θ

⇒ sin θcos θ=512\dfrac{\text{sin θ}}{\text{cos θ}} = \dfrac{5}{12}

⇒ tan θ = 512\dfrac{5}{12}.

⇒(sin θ + cos θ)(2 cos θ - sin θ)[Dividing numerator and denominator by cos θ]⇒(sin θ + cos θ)cos θ(2 cos θ - sin θ)cos θ⇒sin θcos θ+cos θcos θ2 cos θcos θ−sin θcos θ⇒tan θ+12−tan θ [∵tan θ=sin θcos θ]Substituting values we get⇒512+12−512⇒5+121224−512⇒17121912⇒1719.\Rightarrow \dfrac{\text{(sin θ + cos θ)}}{\text{(2 cos θ - sin θ)}} \\[1em] [\text{Dividing numerator and denominator by cos θ}] \\[1em] \Rightarrow \dfrac{\dfrac{\text{(sin θ + cos θ)}}{\text{cos θ}}}{\dfrac{\text{(2 cos θ - sin θ)}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin θ}}{\text{cos θ}} + \dfrac{\text{cos θ}}{\text{cos θ}}}{\dfrac{\text{2 cos θ}}{\text{cos θ}} - \dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\text{tan θ} + 1}{2 - \text{tan θ}} \space [\because \text{tan θ} = \dfrac{\text{sin θ}}{\text{cos θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \dfrac{\dfrac{5}{12} + 1}{2 - \dfrac{5}{12}} \\[1em] \Rightarrow \dfrac{\dfrac{5 + 12}{12}}{\dfrac{24 - 5}{12}} \Rightarrow \dfrac{\dfrac{17}{12}}{\dfrac{19}{12}} \\[1em] \Rightarrow \dfrac{17}{19}.

Hence, (sin θ + cos θ)(2 cos θ - sin θ)=1719\dfrac{\text{(sin θ + cos θ)}}{\text{(2 cos θ - sin θ)}} = \dfrac{17}{19}.

Question 23(ii)

If cosec θ = 1312\dfrac{13}{12}, find the value of 2 sin θ - 3 cos θ4 sin θ - 9 cos θ\dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}}.

Answer

Given,

cosec θ = 1312\dfrac{13}{12}

By formula,

⇒ cot2 θ = cosec2 θ - 1

cot2 θ=(1312)2−1=169144−1=169−144144=25144⇒cot θ=25144=512\text{cot}^2 \text{ θ} = \Big(\dfrac{13}{12}\Big)^2 - 1 \\[1em] = \dfrac{169}{144} - 1 \\[1em] = \dfrac{169 - 144}{144} \\[1em] = \dfrac{25}{144} \\[1em] \Rightarrow \text{cot θ} = \sqrt{\dfrac{25}{144}} = \dfrac{5}{12}

Solving,

⇒2 sin θ - 3 cos θ4 sin θ - 9 cos θ[Dividing numerator and denominator by sin θ]⇒(2 sin θ - 3 cos θ)sin θ(4 sin θ - 9 cos θ)sin θ⇒2 sin θsin θ−3 cos θsin θ4 sin θsin θ−9 cos θsin θ⇒2 - 3 cot θ4−9 cot θ [∵cot θ=cos θsin θ]Substituting values we get⇒2−3×5124−9×512⇒2−544−154⇒8−5416−154⇒3414⇒3.\Rightarrow \dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}} \\[1em] [\text{Dividing numerator and denominator by sin θ}] \\[1em] \Rightarrow \dfrac{\dfrac{\text{(2 sin θ - 3 cos θ)}}{\text{sin θ}}}{\dfrac{\text{(4 sin θ - 9 cos θ)}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{2 sin θ}}{\text{sin θ}} - \dfrac{\text{3 cos θ}}{\text{sin θ}}}{\dfrac{\text{4 sin θ}}{\text{sin θ}} - \dfrac{\text{9 cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{\text{2 - 3 cot θ}}{{4 - \text{9 cot θ}}} \space [\because \text{cot θ} = \dfrac{\text{cos θ}}{\text{sin θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \dfrac{2 - 3 \times \dfrac{5}{12}}{4 - 9\times \dfrac{5}{12}} \\[1em] \Rightarrow \dfrac{2 - \dfrac{5}{4}}{4 - \dfrac{15}{4}} \\[1em] \Rightarrow \dfrac{\dfrac{8 - 5}{4}}{\dfrac{16 - 15}{4}} \\[1em] \Rightarrow \dfrac{\dfrac{3}{4}}{\dfrac{1}{4}} \\[1em] \Rightarrow 3.

Hence, 2 sin θ - 3 cos θ4 sin θ - 9 cos θ=3.\dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}} = 3.

Question 24

If 5 sin θ = 3, find the value of sec θ - tan θsec θ + tan θ\dfrac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}}.

Answer

Given,

5 sin θ = 3

⇒ sin θ = 35\dfrac{3}{5}.

Solving,

⇒sec θ - tan θsec θ + tan θ⇒1cos θ−sin θcos θ1cos θ+sin θcos θ⇒1 - sin θcos θ1 + sin θcos θ⇒1 - sin θ1 + sin θ⇒1−351+35⇒5−355+35⇒2585⇒28⇒14.\Rightarrow \dfrac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos θ}} - \dfrac{\text{sin θ}}{\text{cos θ}}}{\dfrac{1}{\text{cos θ}} + \dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 - sin θ}}{\text{cos θ}}}{\dfrac{\text{1 + sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\text{1 - sin θ}}{\text{1 + sin θ}} \\[1em] \Rightarrow \dfrac{1 - \dfrac{3}{5}}{1 + \dfrac{3}{5}} \\[1em] \Rightarrow \dfrac{\dfrac{5 - 3}{5}}{\dfrac{5 + 3}{5}} \\[1em] \Rightarrow \dfrac{\dfrac{2}{5}}{\dfrac{8}{5}} \\[1em] \Rightarrow \dfrac{2}{8} \\[1em] \Rightarrow \dfrac{1}{4}.

Hence, sec θ - tan θsec θ + tan θ=14\dfrac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} = \dfrac{1}{4}.

Question 25

If θ is an acute angle and sin θ = cos θ, find the value of

2 tan2 θ + sin2 θ - 1.

Answer

Given,

sin θ = cos θ

⇒ sin θcos θ=1\dfrac{\text{sin θ}}{\text{cos θ}} = 1

⇒ tan θ = 1.

⇒ tan θ = tan 45°

⇒ θ = 45°.

⇒ sin 45° = cos 45° = 12\dfrac{1}{\sqrt{2}}

Substituting values in 2 tan2 θ + sin2 θ - 1 we get :

⇒2(1)2+(12)2−1⇒2+12−1⇒1+12⇒2+12⇒32.\Rightarrow 2(1)^2 + \Big(\dfrac{1}{\sqrt{2}}\Big)^2 - 1 \\[1em] \Rightarrow 2 + \dfrac{1}{2} - 1 \\[1em] \Rightarrow 1 + \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{2 + 1}{2} \Rightarrow \dfrac{3}{2}.

Hence, 2 tan2 θ + sin2 θ - 1 = 32\dfrac{3}{2}.

Question 26(i)

Prove the following :

cos θ tan θ = sin θ

Answer

To prove,

cos θ tan θ = sin θ

We know that,

tan θ=sin θcos θ\text{tan θ} = \dfrac{\text{sin θ}}{\text{cos θ}}

Substituting value in L.H.S. of cos θ tan θ = sin θ we get,

cos θ×sin θcos θ=sin θ⇒sin θ=sin θ\text{cos θ} \times \dfrac{\text{sin θ}}{\text{cos θ}} = \text{sin θ} \\[1em] \Rightarrow \text{sin θ} = \text{sin θ}

Since, L.H.S. = R.H.S.

Hence proved that cos θ tan θ = sin θ.

Question 26(ii)

Prove the following :

sin θ cot θ = cos θ

Answer

To prove,

sin θ cot θ = cos θ

We know that,

cot θ = cos θsin θ\dfrac{\text{cos θ}}{\text{sin θ}}

Substituting value in L.H.S. of sin θ cot θ = cos θ we get,

sin θ×cos θsin θ=cos θ⇒cos θ=cos θ\text{sin θ} \times \dfrac{\text{cos θ}}{\text{sin θ}} = \text{cos θ} \\[1em] \Rightarrow \text{cos θ} = \text{cos θ}

Since, L.H.S. = R.H.S.

Hence. proved that sin θ cot θ = cos θ.

Question 26(iii)

Prove the following :

sin2 θcos  θ+cos  θ=1cos  θ\dfrac{\text{sin}^2 \text{ θ}}{\text{cos \text{ θ}}} + \text{cos \text{ θ}} = \dfrac{1}{\text{cos \text{ θ}}}

Answer

To prove,

sin2 θcos  θ+cos  θ=1cos  θ\dfrac{\text{sin}^2 \text{ θ}}{\text{cos \text{ θ}}} + \text{cos \text{ θ}} = \dfrac{1}{\text{cos \text{ θ}}}

Solving LHS of the above equation,

=sin2 θcos θ+cos θ=sin2 θ+cos2θcos θ=1cos θ [∵sin2 θ+ cos2 θ=1]\phantom{=} \dfrac{\text{sin}^2 \text{ θ}}{\text{cos θ}} + \text{cos θ} \\[1em] = \dfrac{\text{sin}^2 \text{ θ} + \text{cos}^2 θ}{\text{cos θ}} \\[1em] = \dfrac{1}{\text{cos θ}} \space [\because \text{sin}^2 \text{ θ} + \text{ cos}^2 \text{ θ} = 1]

Since, L.H.S. = R.H.S.

Hence, proved that sin2 θcos θ+cos θ=1cos θ.\dfrac{\text{sin}^2 \text{ θ}}{\text{cos θ}} + \text{cos θ} = \dfrac{1}{\text{cos θ}}.

Question 27

If in ∆ABC, ∠C = 90° and tan A = 34\dfrac{3}{4}, prove that

sin A cos B + cos A sin B = 1.

Answer

Let ABC be a right angled triangle with ∠C = 90°.

If in ∆ABC, ∠C = 90° and tan A = 3/4, prove that sin A cos B + cos A sin B = 1. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Given,

tan A = 34\dfrac{3}{4}

By formula,

tan A=PerpendicularBase\text{tan A} = \dfrac{\text{Perpendicular}}{\text{Base}}

⇒ 34=BCAC\dfrac{3}{4} = \dfrac{BC}{AC}

Let BC = 3x and AC = 4x.

In △ABC,

⇒ AB2 = AC2 + BC2

⇒ AB2 = (4x)2 + (3x)2

⇒ AB2 = 16x2 + 9x2

⇒ AB2 = 25x2

⇒ AB = 25x2\sqrt{25x^2}

⇒ AB = 5x.

By formula,

sin A=PerpendicularHypotenuse=BCAB=3x5x=35sin B=PerpendicularHypotenuse=ACAB=4x5x=45cos A=BaseHypotenuse=ACAB=4x5x=45cos B=BaseHypotenuse=BCAB=3x5x=35.\text{sin A} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{BC}{AB} = \dfrac{3x}{5x} = \dfrac{3}{5} \\[1em] \text{sin B} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{AC}{AB} = \dfrac{4x}{5x} = \dfrac{4}{5} \\[1em] \text{cos A} = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{AC}{AB} = \dfrac{4x}{5x} = \dfrac{4}{5} \\[1em] \text{cos B} = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{BC}{AB} = \dfrac{3x}{5x} = \dfrac{3}{5}.

Substituting values in L.H.S. of sin A cos B + cos A sin B = 1.

sin A cos B + cos A sin B=35×35+45×45=925+1625=9+1625=2525=1\text{sin A cos B + cos A sin B} = \dfrac{3}{5} \times \dfrac{3}{5} + \dfrac{4}{5} \times \dfrac{4}{5} \\[1em] = \dfrac{9}{25} + \dfrac{16}{25} \\[1em] = \dfrac{9 + 16}{25} \\[1em] = \dfrac{25}{25} \\[1em] = 1

Since, L.H.S. = R.H.S.

Hence, proved that sin A cos B + cos A sin B = 1.

Question 28(a)

In the figure (1) given below, use trigonometry, find DEAD\dfrac{DE}{AD} if BCAB=512\dfrac{BC}{AB} = \dfrac{5}{12}.

In the figure (1) given below, use trigonometry,: Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

By formula, tan θ=PerpendicularBase\text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}}

In ΔABC,

⇒tan θ=BCAB⇒tan θ=512................(1)\Rightarrow \text{tan θ} = \dfrac{BC}{AB} \\[1em] \Rightarrow \text{tan θ} = \dfrac{5}{12}................(1)

In ΔADE,

⇒tan θ=DEAD\Rightarrow \text{tan θ} = \dfrac{DE}{AD}

Substituting the value of tan θ from equation (1), we get

⇒DEAD=512\Rightarrow \dfrac{DE}{AD} = \dfrac{5}{12}.

Hence, DEAD=512\dfrac{DE}{AD} = \dfrac{5}{12}.

Question 28(b)

In the figure (2) given below, ∆ABC is right-angled at B and BD is perpendicular to AC. Find :

(i) cos ∠CBD

(ii) cot ∠ABD

In the figure, ∆ABC is right-angled at B and BD is perpendicular to AC. Find : (i) cos ∠CBD (ii) cot ∠ABD. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In right angle ∆ABC,

⇒ AC2 = AB2 + BC2

⇒ AC2 = 122 + 52

⇒ AC2 = 144 + 25

⇒ AC2 = 169

⇒ AC = 169\sqrt{169} = 13.

Let ∠CBD = x.

∠DBA = 90° - x

In ∆DAB,

⇒ ∠DAB + ∠ADB + ∠DBA = 180° [Angle sum property of triangle]

⇒ ∠DAB + 90° + 90° - x = 180°

⇒ ∠DAB = 180° - 180° + x

⇒ ∠DAB = x.

From figure,

∠DAB = ∠CAB = x

∴ ∠CBD = ∠CAB = x

(i) cos ∠CBD = cos ∠CAB

= ABAC=1213\dfrac{AB}{AC} = \dfrac{12}{13}.

Hence, cos ∠CBD = 1213\dfrac{12}{13}.

(ii) In ∆BCD,

⇒ ∠DBC + ∠DCB + ∠CDB = 180° [Angle sum property of triangle]

⇒ ∠DCB + x + 90° = 180°

⇒ ∠DCB = 180° - 90° - x

⇒ ∠DCB = 90° - x.

From figure,

∠DCB = ∠ACB =90° - x

∴ ∠ABD = ∠ACB = 90° - x

∴ cot ∠ABD = cot ∠ACB

= BCAB=512\dfrac{BC}{AB} = \dfrac{5}{12}.

Hence, cot ∠ABD = 512\dfrac{5}{12}.

Question 29

In the adjoining figure, ABCD is a rectangle. Its diagonal AC = 15 cm and ∠ACD = α. If cot α = 32\dfrac{3}{2}, find the perimeter and the area of the rectangle.

In the figure, ABCD is a rectangle. Its diagonal AC = 15 cm and ∠ACD = α. If cot α = 3/2, find the perimeter and the area of the rectangle. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

Given,

cot α=32\text{cot α} = \dfrac{3}{2} ...........(1)

cot α=BasePerpendicular=CDAD\text{cot α} = \dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{CD}{AD} ........(2)

From (1) and (2) we get,

CDAD=32\dfrac{CD}{AD} = \dfrac{3}{2}

Let CD = 3x and AD = 2x.

In right angle triangle ADC,

⇒AC2=AD2+CD2⇒152=(2x)2+(3x)2⇒225=4x2+9x2⇒13x2=225⇒x2=22513⇒x=22513⇒x=1513.\Rightarrow AC^2 = AD^2 + CD^2 \\[1em] \Rightarrow 15^2 = (2x)^2 + (3x)^2 \\[1em] \Rightarrow 225 = 4x^2 + 9x^2 \\[1em] \Rightarrow 13x^2 = 225 \\[1em] \Rightarrow x^2 = \dfrac{225}{13} \\[1em] \Rightarrow x = \sqrt{\dfrac{225}{13}} \\[1em] \Rightarrow x = \dfrac{15}{\sqrt{13}}.

Perimeter of rectangle (P) = 2(CD + AD)

Area of rectangle (A) = CD × AD

Substituting values we get :

P=2×(3x+2x)=2×5x=10x=10×1513 cm.=15013 cm.A=3x×2x=6x2=6×(1513)2=6×22513=135013=1031113 cm2.P = 2 \times (3x + 2x) \\[1em] = 2 \times 5x \\[1em] = 10x \\[1em] = 10 \times \dfrac{15}{\sqrt{13}} \text{ cm}. \\[1em] = \dfrac{150}{\sqrt{13}} \text{ cm}. \\[1em] A = 3x \times 2x \\[1em] = 6x^2 \\[1em] = 6 \times \Big(\dfrac{15}{\sqrt{13}}\Big)^2 \\[1em] = 6 \times \dfrac{225}{13} \\[1em] = \dfrac{1350}{13} \\[1em] = 103\dfrac{11}{13} \text{ cm}^2.

Hence, perimeter of rectangle = 15013\dfrac{150}{\sqrt{13}} cm and area = 1031113103\dfrac{11}{13} cm2.

Question 30

Using the measurements given in the figure alongside,

(a) Find the values of:

(i) sin Φ

(ii) tan θ.

(b) Write an expression for AD in terms of θ.

Using the measurements given in the figure, (a) Find the values of: (i) sin Φ (ii) tan θ. (b) Write an expression for AD in terms of θ. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

(a) In right-angled ∆BCD

Using pythagoras theorem we get :

⇒ BD2 = BC2 + CD2

⇒ CD2 = BD2 - BC2

⇒ CD2 = (13)2 - (12)2

⇒ CD2 = 169 - 144

⇒ CD2 = 25

⇒ CD = 25\sqrt{25} = 5.

(i) By formula,

sin Φ = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

In right-angled ∆BCD

sin Φ = CDBD=513\dfrac{CD}{BD} = \dfrac{5}{13}.

Hence, sin Φ = 513\dfrac{5}{13}.

(ii) Draw DE perpendicular to AB.

Using the measurements given in the figure, (a) Find the values of: (i) sin Φ (ii) tan θ. (b) Write an expression for AD in terms of θ. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

From figure,

ED = BC = 12

In right-angled ∆BED

Using pythagoras theorem we get :

⇒ BD2 = ED2 + EB2

⇒ 132 = 122 + EB2

⇒ EB2 = (13)2 - (12)2

⇒ EB2 = 169 - 144

⇒ EB2 = 25

⇒ EB = 25\sqrt{25} = 5.

From figure,

AE = AB - EB = 14 - 5 = 9.

By formula,

tan θ = PerpendicularBase=EDAE=129=43\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{ED}{AE} = \dfrac{12}{9} = \dfrac{4}{3}.

Hence, tan θ = 43\dfrac{4}{3}.

(b) In right-angled ∆AED

⇒sin θ=PerpendicularHypotenuse⇒sin θ=EDAD⇒sin θ=12AD⇒AD=12sin θ.⇒cos θ=BaseHypotenuse⇒cos θ=AEAD⇒cos θ=9AD⇒AD=9cos θ.\phantom{\Rightarrow} \text{sin θ} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow \text{sin θ} = \dfrac{ED}{AD} \\[1em] \Rightarrow \text{sin θ} = \dfrac{12}{AD} \\[1em] \Rightarrow AD = \dfrac{12}{\text{sin θ}}. \\[1em] \phantom{\Rightarrow} \text{cos θ} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow \text{cos θ} = \dfrac{AE}{AD} \\[1em] \Rightarrow \text{cos θ} = \dfrac{9}{AD} \\[1em] \Rightarrow AD = \dfrac{9}{\text{cos θ}}.

Hence, AD = 9cos θ or 12sin θ\dfrac{9}{\text{cos θ}} \text{ or } \dfrac{12}{\text{sin θ}}.

Question 31(i)

Prove the following:

(sin A + cos A)2 + (sin A - cos A)2 = 2

Answer

Solving L.H.S. of equation : (sin A + cos A)2 + (sin A - cos A)2 = 2, we get,

⇒ (sin A + cos A)2 + (sin A - cos A)2

⇒ sin2 A + cos2 A + 2 sin A cos A + sin2 A + cos2 A - 2 sin A cos A

Since, sin2 A + cos2 A = 1.

⇒ 1 + 2 sin A cos A + 1 - 2 sin A cos A

⇒ 2.

Since, L.H.S. = R.H.S.

Hence, proved that (sin A + cos A)2 + (sin A - cos A)2 = 2.

Question 31(ii)

Prove the following:

cot2 A - 1sin2A\dfrac{1}{\text{sin}^2 A} + 1 = 0

Answer

Solving L.H.S. of the equation : cot2 A - 1sin2A\dfrac{1}{\text{sin}^2 A} + 1 = 0

⇒cos2Asin2A−1sin2A+1⇒cos2A−1+sin2Asin2A⇒sin2A+cos2A−1sin2A⇒1−1sin2A⇒0sin2A⇒0.\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin}^2 A} - \dfrac{1}{\text{sin}^2 A} + 1 \\[1em] \Rightarrow \dfrac{\text{cos}^2 A - 1 + \text{sin}^2 A}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A - 1}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{1 - 1}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{0}{\text{sin}^2 A} \\[1em] \Rightarrow 0.

Since, L.H.S. = R.H.S.

Hence, proved that cot2A−1sin2A+1=0\text{cot}^2 A - \dfrac{1}{\text{sin}^2 A} + 1 = 0

Question 31(iii)

Prove the following:

11 + tan2A+11 + cot2A\dfrac{1}{\text{1 + tan}^2 A} + \dfrac{1}{\text{1 + cot}^2 A} = 1

Answer

Solving L.H.S. of the equation : 11 + tan2A+11 + cot2A\dfrac{1}{\text{1 + tan}^2 A} + \dfrac{1}{\text{1 + cot}^2 A} = 1.

=11+sin2Acos2A+11+cos2Asin2A=1cos2A+sin2Acos2A+1sin2A+cos2Asin2A=cos2Asin2A+cos2A+sin2Asin2A+cos2A=sin2A+cos2Asin2A+cos2A=1.\phantom{=} \dfrac{1}{1 + \dfrac{\text{sin}^2 A}{\text{cos}^2 A}} + \dfrac{1}{1 + \dfrac{\text{cos}^2 A}{\text{sin}^2 A}} \\[1em] = \dfrac{1}{\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos}^2 A}} + \dfrac{1}{\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A}} \\[1em] = \dfrac{\text{cos}^2 A}{\text{sin}^2 A + \text{cos}^2 A} + \dfrac{\text{sin}^2 A}{\text{sin}^2 A + \text{cos}^2 A} \\[1em] = \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A + \text{cos}^2 A} \\[1em] = 1.

Since, L.H.S. = R.H.S.

Hence, proved that 11 + tan2A+11 + cot2A\dfrac{1}{\text{1 + tan}^2 A} + \dfrac{1}{\text{1 + cot}^2 A} = 1.

Question 32

Simplify 1−sin2 θ1−cos2 θ\sqrt{\dfrac{1 - \text{sin}^2 \text{ θ}}{1 - \text{cos}^2 \text{ θ}}}.

Answer

Substituting, 1 - sin2 θ = cos2 θ and 1 - cos2 θ = sin2 θ in 1−sin2 θ1−cos2 θ\sqrt{\dfrac{1 - \text{sin}^2 \text{ θ}}{1 - \text{cos}^2 \text{ θ}}} we get :

⇒cos2 θsin2 θ⇒cot2 θ⇒cot θ.\Rightarrow \sqrt{\dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}}} \\[1em] \Rightarrow \sqrt{\text{cot}^2 \text{ θ}} \\[1em] \Rightarrow \text{cot θ}.

Hence, 1−sin2 θ1−cos2 θ\sqrt{\dfrac{1 - \text{sin}^2 \text{ θ}}{1 - \text{cos}^2 \text{ θ}}} = cot θ.

Question 33

If sin θ + cosec θ = 2, find the value of sin2 θ + cosec2 θ.

Answer

Given,

⇒ sin θ + cosec θ = 2

Squaring both sides we get :

⇒ (sin θ + cosec θ)2 = 22

⇒ sin2 θ + cosec2 θ + 2 sin θ. cosec θ = 4

⇒ sin2 θ + cosec2 θ + 2 sin θ×1sin θ=42\text{ sin θ} \times \dfrac{1}{\text{sin θ}} = 4

⇒ sin2 θ + cosec2 θ + 2 = 4

⇒ sin2 θ + cosec2 θ = 4 - 2

⇒ sin2 θ + cosec2 θ = 2.

Hence, proved that sin2 θ + cosec2 θ = 2.

Question 34

If x = a cos θ + b sin θ and y = a sin θ - b cos θ, prove that x2 + y2 = a2 + b2.

Answer

x2 + y2 = (a cos θ + b sin θ)2 + (a sin θ - b cos θ)2

⇒ x2 + y2 = a2 cos2 θ + b2 sin2 θ + 2ab cos θ. sin θ + a2 sin2 θ + b2 cos2 θ - 2ab cos θ. sin θ

⇒ x2 + y2 = a2(sin2 θ + cos2 θ) + b2(sin2 θ + cos2 θ)

⇒ x2 + y2 = (a2 + b2)(sin2 θ + cos2 θ)

As, sin2 θ + cos2 θ = 1.

⇒ x2 + y2 = (a2 + b2).

Hence proved that x2 + y2 = (a2 + b2).

Multiple Choice Questions

Question 1

In the adjoining figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. Using the figure answer the question.

In the figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. The value of sin A is? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

The value of sin A is

  1. 724\dfrac{7}{24}

  2. 725\dfrac{7}{25}

  3. 257\dfrac{25}{7}

  4. 2425\dfrac{24}{25}

Answer

In right angled triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ AC2 = (24)2 + (7)2

⇒ AC2 = 576 + 49

⇒ AC2 = 625

⇒ AC = 625\sqrt{625} = 25 cm.

By formula,

sin A = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= BCAC=725\dfrac{BC}{AC} = \dfrac{7}{25}.

Hence, Option 2 is the correct option.

Question 2

In the adjoining figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. Using the figure answer the question.

In the figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. The value of sec A is? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

The value of sec A is

  1. 247\dfrac{24}{7}

  2. 724\dfrac{7}{24}

  3. 2524\dfrac{25}{24}

  4. 257\dfrac{25}{7}

Answer

By formula,

sec A = HypotenuseBase\dfrac{\text{Hypotenuse}}{\text{Base}}

= ACAB=2524\dfrac{AC}{AB} = \dfrac{25}{24}.

Hence, Option 3 is the correct option.

Question 3

In the adjoining figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. Using the figure answer the question.

In the figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. The value of tan C is? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

The value of tan C is

  1. 247\dfrac{24}{7}

  2. 724\dfrac{7}{24}

  3. 725\dfrac{7}{25}

  4. 2425\dfrac{24}{25}

Answer

By formula,

tan C = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= ABBC=247\dfrac{AB}{BC} = \dfrac{24}{7}.

Hence, Option 1 is the correct option.

Question 4

In the adjoining figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. Using the figure answer the question.

In the figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. The value of cosec C is? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

The value of cosec C is

  1. 724\dfrac{7}{24}

  2. 2425\dfrac{24}{25}

  3. 257\dfrac{25}{7}

  4. 2524\dfrac{25}{24}

Answer

By formula,

cosec C = HypotenusePerpendicular\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}

= ACAB=2524\dfrac{AC}{AB} = \dfrac{25}{24}.

Hence, Option 4 is the correct option.

Question 5

In the adjoining figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. Using the figure answer the question.

In the figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. The value of tan A + cot C is? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

The value of tan A + cot C is

  1. 712\dfrac{7}{12}

  2. 127\dfrac{12}{7}

  3. 1425\dfrac{14}{25}

  4. 2512\dfrac{25}{12}

Answer

By formula,

tan A = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= BCAB=724\dfrac{BC}{AB} = \dfrac{7}{24}.

cot C = BasePerpendicular\dfrac{\text{Base}}{\text{Perpendicular}}

= BCAB=724\dfrac{BC}{AB} = \dfrac{7}{24}.

Substituting value in tan A + cot C we get :

tan A + cot C=724+724=1424=712.\text{tan A + cot C} = \dfrac{7}{24} + \dfrac{7}{24} \\[1em] = \dfrac{14}{24} \\[1em] = \dfrac{7}{12}.

Hence, Option 1 is the correct option.

Question 6

In the adjoining figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. Using the figure answer the question.

In the figure, ABC is a right angled triangle right angled at B; AB = 24 cm and BC = 7 cm. The value of 2 cos A - sin C is? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

The value of 2 cos A - sin C is

  1. 2524\dfrac{25}{24}

  2. 2425\dfrac{24}{25}

  3. 4125\dfrac{41}{25}

  4. 4925\dfrac{49}{25}

Answer

By formula,

cos A = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= ABAC=2425\dfrac{AB}{AC} = \dfrac{24}{25}.

sin C = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= ABAC=2425\dfrac{AB}{AC} = \dfrac{24}{25}.

Substituting value in 2 cos A - sin C we get :

2 cos A - sin C=2×2425−2425=4825−2425=2425.\text{2 cos A - sin C} = 2 \times \dfrac{24}{25} - \dfrac{24}{25} \\[1em] = \dfrac{48}{25} - \dfrac{24}{25} \\[1em] = \dfrac{24}{25}.

Hence, Option 2 is the correct option.

Question 7

In the adjoining figure, the value of sin B cos C + sin C cos B is

  1. 0

  2. 1

  3. 53\dfrac{5}{3}

  4. 2

In the figure, the value of sin B cos C + sin C cos B is? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In right angle triangle ABC,

⇒ BC2 = AB2 + AC2

⇒ 102 = (6)2 + (AC)2

⇒ AC2 = 102 - 62

⇒ AC2 = 100 - 36

⇒ AC2 = 64

⇒ AC = 64\sqrt{64} = 8 cm.

By formula,

sin B = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= ACBC=810\dfrac{AC}{BC} = \dfrac{8}{10}.

sin C = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= ABBC=610\dfrac{AB}{BC} = \dfrac{6}{10}.

cos B = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= ABBC=610\dfrac{AB}{BC} = \dfrac{6}{10}.

cos C = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= ACBC=810\dfrac{AC}{BC} = \dfrac{8}{10}.

Substituting values in sin B cos C + sin C cos B we get :

⇒sin B cos C + sin C cos B=810×810+610×610=64100+36100=100100=1.\Rightarrow \text{sin B cos C + sin C cos B} = \dfrac{8}{10} \times \dfrac{8}{10} + \dfrac{6}{10} \times \dfrac{6}{10} \\[1em] = \dfrac{64}{100} + \dfrac{36}{100} \\[1em] = \dfrac{100}{100} = 1.

Hence, Option 2 is the correct option.

Question 8

In the adjoining figure, the value of cos θ is

  1. 1213\dfrac{12}{13}

  2. 1312\dfrac{13}{12}

  3. 512\dfrac{5}{12}

  4. 513\dfrac{5}{13}

In the figure, the value of cos θ is? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In right angled triangle BDC,

⇒ BC2 = BD2 + CD2

⇒ BC2 = (4)2 + (3)2

⇒ BC2 = 16 + 9

⇒ BC2 = 25

⇒ BC = 25\sqrt{25} = 5 cm.

In right angled triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ AC2 = (12)2 + (5)2

⇒ AC2 = 144 + 25

⇒ AC2 = 169

⇒ AC = 169\sqrt{169} = 13 cm.

By formula,

cos θ = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= ABAC=1213\dfrac{AB}{AC} = \dfrac{12}{13}.

Hence, Option 1 is the correct option.

Question 9

If cos A = 45\dfrac{4}{5}, then the value of tan A is

  1. 35\dfrac{3}{5}

  2. 34\dfrac{3}{4}

  3. 43\dfrac{4}{3}

  4. 53\dfrac{5}{3}

Answer

Let ABC be a right angle triangle with ∠B = 90°.

If cos A = 4/5, then the value of tan A is? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By formula,

cos A = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

Substituting values we get :

⇒45=ABAC\Rightarrow \dfrac{4}{5} = \dfrac{AB}{AC}

Let AB = 4x and AC = 5x.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ (5x)2 = (4x)2 + BC2

⇒ 25x2 = 16x2 + BC2

⇒ BC2 = 25x2 - 16x2

⇒ BC2 = 9x2

⇒ BC = 9x2\sqrt{9x^2} = 3x.

By formula,

tan A = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

= BCAB=3x4x=34\dfrac{BC}{AB} = \dfrac{3x}{4x} = \dfrac{3}{4}.

Hence, Option 2 is the correct option.

Question 10

If sin A = 12\dfrac{1}{2}, then the value of cot A is

  1. 3\sqrt{3}

  2. 13\dfrac{1}{\sqrt{3}}

  3. 32\dfrac{\sqrt{3}}{2}

  4. 1

Answer

Let ABC be a right angle triangle with ∠B = 90°.

If sin A = 1/2, then the value of cot A is? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By formula,

sin A = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

Substituting values we get :

⇒12=BCAC\Rightarrow \dfrac{1}{2} = \dfrac{BC}{AC}

Let BC = x and AC = 2x.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ (2x)2 = AB2 + (x)2

⇒ 4x2 = AB2 + x2

⇒ AB2 = 3x2

⇒ AB = 3x2=3x\sqrt{3x^2} = \sqrt{3}x.

By formula,

cot A = BasePerpendicular=ABBC=3xx=3\dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{AB}{BC} = \dfrac{\sqrt{3}x}{x} = \sqrt{3}.

Hence, Option 1 is the correct option.

Question 11

If cosec θ = 1312\dfrac{13}{12}, then the value of tan θ is

  1. 125\dfrac{12}{5}

  2. 512\dfrac{5}{12}

  3. 513\dfrac{5}{13}

  4. 512\dfrac{5}{12}

Answer

Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.

If cosec θ = 13/12, then the value of tan θ is? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By formula,

cosec θ = HypotenusePerpendicular\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}

Substituting values we get :

⇒1312=ACAB\Rightarrow \dfrac{13}{12} = \dfrac{AC}{AB}

Let AC = 13x and AB = 12x.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ (13x)2 = (12x)2 + BC2

⇒ 169x2 = 144x2 + BC2

⇒ BC2 = 169x2 - 144x2

⇒ BC2 = 25x2

⇒ BC = 25x2=5x\sqrt{25x^2} = 5x.

By formula,

tan θ = PerpendicularBase=ABBC=12x5x=125\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{BC} = \dfrac{12x}{5x} = \dfrac{12}{5}.

Hence, Option 1 is the correct option.

Question 12

If tan A = xy\dfrac{x}{y}, then cos A is equal to

  1. xx2+y2\dfrac{x}{\sqrt{x^2 + y^2}}

  2. yx2+y2\dfrac{y}{\sqrt{x^2 + y^2}}

  3. x2−y2x2+y2\dfrac{x^2 - y^2}{\sqrt{x^2 + y^2}}

  4. x2−y2x2+y2\dfrac{x^2 - y^2}{x^2 + y^2}

Answer

Let ABC be a right angle triangle with ∠B = 90°.

If tan A = x/y, then cos A is equal to? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By formula,

tan A = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

Substituting values we get :

⇒xy=BCAB\Rightarrow \dfrac{x}{y} = \dfrac{BC}{AB}

Let BC = xk and AB = yk.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ AC2 = (yk)2 + (xk)2

⇒ AC2 = y2k2 + x2k2

⇒ AC2 = k2(y2 + x2)

⇒ AC = k2(y2+x2)\sqrt{k^2(y^2 + x^2)}

⇒ AC = ky2+x2k\sqrt{y^2 + x^2}.

By formula,

cos A = BaseHypotenuse=ABAC=ykkx2+y2=yx2+y2\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{yk}{k\sqrt{x^2 + y^2}} = \dfrac{y}{\sqrt{x^2 + y^2}}.

Hence, Option 2 is the correct option.

Question 13

If sin θ = ab\dfrac{a}{b}, then cos θ is equal to

  1. bb2−a2\dfrac{b}{\sqrt{b^2 - a^2}}

  2. ba\dfrac{b}{a}

  3. b2−a2b\dfrac{\sqrt{b^2 - a^2}}{b}

  4. ab2−a2\dfrac{a}{\sqrt{b^2 - a^2}}

Answer

Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.

If sin θ = a/b, then cos θ is equal to? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By formula,

sin θ = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

Substituting values we get :

⇒ab=ABAC\Rightarrow \dfrac{a}{b} = \dfrac{AB}{AC}

Let AB = ak and AC = bk.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ (bk)2 = (ak)2 + BC2

⇒ b2k2 = a2k2 + BC2

⇒ BC2 = b2k2 - a2k2

⇒ BC2 = k2(b2−a2)\sqrt{k^2(b^2 - a^2)}

⇒ BC = k(b2−a2)k\sqrt{(b^2 - a^2)}.

By formula,

cos θ=BaseHypotenuse=BCAC=k(b2−a2)bk=b2−a2b\text{cos θ} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{BC}{AC} \\[1em] = \dfrac{k\sqrt{(b^2 - a^2)}}{bk} \\[1em] = \dfrac{\sqrt{b^2 - a^2}}{b}

Hence, Option 3 is the correct option.

Question 14

Consider the following two statements:

Statement 1: In sin A = 12\dfrac{1}{2}, then value of cot A is 13\dfrac{1}{\sqrt{3}}.

Statement 2: cot A = sin A.cos A.

Which of the following is valid?

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and Statement 2 is false.

  4. Statement 1 is false, and Statement 2 is true.

Answer

Given, sin A = 12\dfrac{1}{2}

By formula,

⇒ sin2 A + cos2 A = 1

⇒ (12)2\Big(\dfrac{1}{2}\Big)^2 + cos2 A = 1

⇒ 14\dfrac{1}{4} + cos2 A = 1

⇒ cos2 A = 1−141 - \dfrac{1}{4}

⇒ cos2 A = 34\dfrac{3}{4}

⇒ cos A = 34=32\sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{2}.

By formula,

⇒ cot A = cos Asin A=3212=232=3\dfrac{\text{cos A}}{\text{sin A}} = \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \dfrac{2\sqrt{3}}{2} = \sqrt{3}.

Thus, both the statements are false.

Hence, option 2 is the correct option.

Assertion Reason Type Questions

Question 1

Assertion (A): In the adjoining figure, tan A = 13\dfrac{1}{\sqrt{3}}. Then AC = 2AB.

Reason (R): In right angled ΔABC, AC2 = AB2 + BC2

In the adjoining figure, tan A = 1/3. Then AC = 2AB. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

  1. Assertion (A) is true, Reason (R) is false.

  2. Assertion (A) is false, Reason (R) is true.

  3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

  4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

Given, tan A = 13\dfrac{1}{\sqrt{3}}

As we know,

tan A =PerpendicularBase⇒BCAB=13\text{tan A }= \dfrac{\text{Perpendicular}}{\text{Base}}\\[1em] \Rightarrow \dfrac{BC}{AB} = \dfrac{1}{\sqrt{3}}

Let BC = k and AB = 3\sqrt{3} k

Since, ΔABC is a right angled triangle, using pythagoras theorem,

⇒ AC2 = AB2 + BC2

∴ Reason (R) is true.

Substituting values we get :

⇒AC2=(3k)2+k2⇒AC2=3k2+k2⇒AC2=4k2⇒AC=4k2⇒AC=2k⇒AC=2k33⇒AC=2k×33⇒AC=2(3k)3⇒AC=2AB3.\Rightarrow AC^2 = (\sqrt{3}k)^2 + k^2\\[1em] \Rightarrow AC^2 = 3k^2 + k^2\\[1em] \Rightarrow AC^2 = 4k^2\\[1em] \Rightarrow AC = \sqrt{4k^2}\\[1em] \Rightarrow AC = 2k \\[1em] \Rightarrow AC = 2k\dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow AC = 2k \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{2(\sqrt{3}k)}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{2AB}{\sqrt{3}}.

∴ Assertion (A) is false.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Question 2

Assertion (A): In adjoining triangle ABC, sinA cosA = 1225\dfrac{12}{25}.

Reason (R): cos A = 1sin A\dfrac{1}{\text{sin A}}.

In adjoining triangle ABC, sinA cosA =: Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

  1. Assertion (A) is true, Reason (R) is false.

  2. Assertion (A) is false, Reason (R) is true.

  3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

  4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

In triangle ABC,

Using pythagoras theorem,

⇒ AC2 = AB2 + BC2

⇒ AC2 = 42 + 32

⇒ AC2 = 16 + 9

⇒ AC2 = 25

⇒ AC = 25\sqrt{25}

⇒ AC = 5

By formula,

sin A=PerpendicularHypotenuse=BCAC=35.\text{sin A} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}\\[1em] = \dfrac{BC}{AC}\\[1em] = \dfrac{3}{5}.

By formula,

cos A=BaseHypotenuse=ABAC=45.\text{cos A} = \dfrac{\text{Base}}{\text{Hypotenuse}}\\[1em] = \dfrac{AB}{AC}\\[1em] = \dfrac{4}{5}.

Substituting values we get,

sin A cos A = 35×45=3×45×5=1225\dfrac{3}{5} \times \dfrac{4}{5} = \dfrac{3 \times 4}{5 \times 5} = \dfrac{12}{25}.

∴ Assertion (A) is true.

By formula,

1sin A\dfrac{1}{\text{sin A}} = cosec A ≠ cos A

∴ Reason (R) is false.

∴ Assertion (A) is true, Reason (R) is false.

Hence, option 1 is the correct option.

Question 3

Assertion (A): If x = a cos θ + b sin θ and y = a cos θ - b sin θ, then x2 + y2 = a2 + b2

Reason (R): cos2θ + sin2θ = 1.

  1. Assertion (A) is true, Reason (R) is false.

  2. Assertion (A) is false, Reason (R) is true.

  3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

  4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

Given, x = a cos θ + b sin θ and y = a cos θ - b sin θ

⇒ x2 = (a cos θ + b sin θ)2

⇒ x2 = a2 cos2 θ + b2 sin2 θ + 2 ab cos θ sin θ .....................(1)

Similarly,

⇒ y2 = (a cos θ - b sin θ)2

⇒ y2 = a2 cos2 θ + b2 sin2 θ - 2 a b cos θ sin θ .....................(2)

Adding equations (1) and (2), we get :

⇒ x2 + y2 = (a2 cos2 θ + b2 sin2 θ + 2ab cos θ sin θ) + (a2 cos2 θ + b2 sin2 θ - 2ab cos θ sin θ)

⇒ x2 + y2 = (a2 cos2 θ + a2 cos2 θ) + (b2 sin2 θ + b2 sin2 θ) + (2ab cos θ sin θ - 2ab cos θ sin θ)

⇒ x2 + y2 = 2a2 cos2 θ + 2 b2 sin2 θ

⇒ x2 + y2 = 2(a2 cos2 θ + b2 sin2 θ)

∴ Assertion (A) is false.

cos2 θ + sin2 θ = 1

This is a fundamental Pythagorean trigonometric identity and is always true for any real value of θ.

∴ Reason (R) is true.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Chapter Test

Question 1(a)

From the figure (i) given below, calculate all the six t-ratios for both acute angles.

From the figure, calculate all the six t-ratios for both acute angles. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In right angle triangle ABC,

By pythagoras theorem we get :

⇒ AC2 = AB2 + BC2

⇒ 32 = AB2 + 22

⇒ 9 = AB2 + 4

⇒ AB2 = 9 - 4

⇒ AB2 = 5

⇒ AB = 5\sqrt{5}.

For angle A,

⇒sin A=PerpendicularHypotenuse=BCAC=23.⇒cos A=BaseHypotenuse=ABAC=53.⇒tan A=PerpendicularBase=BCAB=25.⇒cot A=BasePerpendicular=ABBC=52.⇒sec A=HypotenuseBase=ACAB=35.⇒cosec A=HypotenusePerpendicular=ACBC=32.\Rightarrow \text{sin A} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{BC}{AC} = \dfrac{2}{3}. \\[1em] \Rightarrow \text{cos A} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{AB}{AC} = \dfrac{\sqrt{5}}{3}. \\[1em] \Rightarrow \text{tan A} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{BC}{AB} = \dfrac{2}{\sqrt{5}}. \\[1em] \Rightarrow \text{cot A} = \dfrac{\text{Base}}{\text{Perpendicular}} \\[1em] = \dfrac{AB}{BC} = \dfrac{\sqrt{5}}{2}. \\[1em] \Rightarrow \text{sec A} = \dfrac{\text{Hypotenuse}}{\text{Base}} \\[1em] = \dfrac{AC}{AB} = \dfrac{3}{\sqrt{5}}. \\[1em] \Rightarrow \text{cosec A} = \dfrac{\text{Hypotenuse}}{\text{Perpendicular}} \\[1em] = \dfrac{AC}{BC} = \dfrac{3}{2}. \\[1em]

For angle C,

⇒sin C=PerpendicularHypotenuse=ABAC=53.⇒cos C=BaseHypotenuse=BCAC=23.⇒tan C=PerpendicularBase=ABBC=52.⇒cot C=BasePerpendicular=BCAB=25.⇒sec C=HypotenuseBase=ACBC=32.⇒cosec C=HypotenusePerpendicular=ACAB=35.\Rightarrow \text{sin C} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{AB}{AC} = \dfrac{\sqrt{5}}{3}. \\[1em] \Rightarrow \text{cos C} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{BC}{AC} = \dfrac{2}{3}. \\[1em] \Rightarrow \text{tan C} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AB}{BC} = \dfrac{\sqrt{5}}{2}. \\[1em] \Rightarrow \text{cot C} = \dfrac{\text{Base}}{\text{Perpendicular}} \\[1em] = \dfrac{BC}{AB} = \dfrac{2}{\sqrt{5}}. \\[1em] \Rightarrow \text{sec C} = \dfrac{\text{Hypotenuse}}{\text{Base}} \\[1em] = \dfrac{AC}{BC} = \dfrac{3}{2}. \\[1em] \Rightarrow \text{cosec C} = \dfrac{\text{Hypotenuse}}{\text{Perpendicular}} \\[1em] = \dfrac{AC}{AB} = \dfrac{3}{\sqrt{5}}. \\[1em]

Question 1(b)

From the figure (ii) given below, find the values of x and y in terms of t-ratios of θ.

From the figure, find the values of x and y in terms of t-ratios of θ. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

From figure,

From the figure, find the values of x and y in terms of t-ratios of θ. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

⇒cot θ=BasePerpendicular⇒cot θ=ABBC⇒cot θ=x10⇒x=10 cot θ.⇒cosec θ=HypotenusePerpendicular⇒cosec θ=ACBC⇒cosec θ=y10⇒y=10 cosec θ.\Rightarrow \text{cot θ} = \dfrac{\text{Base}}{\text{Perpendicular}} \\[1em] \Rightarrow \text{cot θ} = \dfrac{AB}{BC} \\[1em] \Rightarrow \text{cot θ} = \dfrac{x}{10} \\[1em] \Rightarrow x = 10 \text{ cot θ}. \\[1em] \Rightarrow \text{cosec θ} = \dfrac{\text{Hypotenuse}}{\text{Perpendicular}} \\[1em] \Rightarrow \text{cosec θ} = \dfrac{AC}{BC} \\[1em] \Rightarrow \text{cosec θ} = \dfrac{y}{10} \\[1em] \Rightarrow y = 10 \text{ cosec θ}.

Hence, x = 10 cot θ and y = 10 cosec θ.

Question 2(a)

From the figure (1) given below, find the values of :

(i) sin ∠ABC

(ii) tan x - cos x + 3 sin x

From the figure, find the values of (i) sin ∠ABC (ii) tan x - cos x + 3 sin x. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

(i) In right angle triangle ABC,

By pythagoras theorem we get :

⇒ AB2 = AC2 + BC2

⇒ 202 = AC2 + 122

⇒ 400 = AC2 + 144

⇒ AC2 = 400 - 144

⇒ AC2 = 256

⇒ AC = 256\sqrt{256} = 16.

sin ∠ABC=PerpendicularHypotenuse=ACAB=1620=45.\text{sin ∠ABC} = \dfrac{\text{Perpendicular}}{\text{\text{Hypotenuse}}} \\[1em] = \dfrac{AC}{AB} \\[1em] = \dfrac{16}{20} = \dfrac{4}{5}.

Hence, sin ∠ABC = 45\dfrac{4}{5}.

(ii) In right angle triangle BCD,

By pythagoras theorem we get :

⇒ BD2 = BC2 + CD2

⇒ BD2 = 122 + 92

⇒ BD2 = 144 + 81

⇒ BD2 = 225

⇒ BD = 225\sqrt{225}

⇒ BD = 15.

By formula,

tan x =PerpendicularBase=BCCD=129=43.cos x =BaseHypotenuse=CDBD=915=35.sin x =PerpendicularHypotenuse=BCBD=1215=45.\text{tan x } = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{BC}{CD} = \dfrac{12}{9} = \dfrac{4}{3}. \\[1em] \text{cos x } = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{CD}{BD} = \dfrac{9}{15} = \dfrac{3}{5}. \\[1em] \text{sin x } = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{BC}{BD} = \dfrac{12}{15} = \dfrac{4}{5}. \\[1em]

Substituting values in tan x - cos x + 3 sin x we get :

⇒43−35+3×45⇒43−35+125⇒20−9+3615⇒4715⇒3215.\Rightarrow \dfrac{4}{3} - \dfrac{3}{5} + 3 \times \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{4}{3} - \dfrac{3}{5} + \dfrac{12}{5} \\[1em] \Rightarrow \dfrac{20 - 9 + 36}{15} \\[1em] \Rightarrow \dfrac{47}{15} \\[1em] \Rightarrow 3\dfrac{2}{15}.

Hence, tan x - cos x + 3 sin x = 3215.3\dfrac{2}{15}.

Question 2(b)

From the figure (2) given below, find the values of :

(i) 5 sin x

(ii) 7 tan x

(iii) 5 cos x - 17 sin y - tan x

From the figure, find the values of (i) 5 sin x (ii) 7 tan x (iii) 5 cos x - 17 sin y - tan x. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

(i) By formula,

sin x =PerpendicularHypotenuse=ADAB=1525=35.5 sin x=5×35=3.\text{sin x } = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{AD}{AB} = \dfrac{15}{25} = \dfrac{3}{5}. \\[1em] 5\text{ sin x} = 5 \times \dfrac{3}{5} = 3.

Hence, 5 sin x = 3.

(ii) In right angle triangle ABD,

⇒ AB2 = AD2 + BD2

⇒ 252 = 152 + BD2

⇒ 625 = 225 + BD2

⇒ BD2 = 625 - 225

⇒ BD2 = 400

⇒ BD = 400\sqrt{400} = 20.

By formula,

tan x =PerpendicularBase=ADBD=1520=34.7 tan x=7×34=214=514.\text{tan x } = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AD}{BD} = \dfrac{15}{20} = \dfrac{3}{4}. \\[1em] 7\text{ tan x} = 7 \times \dfrac{3}{4} = \dfrac{21}{4} = 5\dfrac{1}{4}.

Hence, 7 tan x = 5145\dfrac{1}{4}.

(iii) In right angle triangle ADC,

⇒ AC2 = AD2 + DC2

⇒ 172 = 152 + DC2

⇒ 289 = 225 + DC2

⇒ DC2 = 289 - 225

⇒ DC2 = 64

⇒ DC = 64\sqrt{64} = 8.

Solving,

5 cos x - 17 sin y - tan x=5×BDAB−17×CDAC−ADBD=5×2025−17×817−1520=4−8−34=−4−34=−16−34=−194=−434.\text{5 cos x - 17 sin y - tan x} = 5 \times \dfrac{BD}{AB} - 17 \times \dfrac{CD}{AC} - \dfrac{AD}{BD} \\[1em] = 5 \times \dfrac{20}{25} - 17 \times \dfrac{8}{17} - \dfrac{15}{20} \\[1em] = 4 - 8 - \dfrac{3}{4} \\[1em] = -4 - \dfrac{3}{4} \\[1em] = \dfrac{-16 - 3}{4}\\[1em] = -\dfrac{19}{4} \\[1em] = -4\dfrac{3}{4}.

Hence, 5 cos x - 17 sin y - tan x = -434.4\dfrac{3}{4}.

Question 3

If q cos θ = p, find tan θ - cot θ in terms of p and q.

Answer

Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.

If q cos θ = p, find tan θ - cot θ in terms of p and q. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Given,

⇒ q cos θ = p

⇒ cos θ = pq\dfrac{p}{q} ..........(1)

By formula,

⇒ cos θ = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

⇒ cos θ = BCAC\dfrac{BC}{AC} ..........(2)

Comparing equations (1) and (2) we get :

⇒pq=BCAC\Rightarrow \dfrac{p}{q} = \dfrac{BC}{AC}

Let BC = pk and AC = qk.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ (qk)2 = AB2 + (pk)2

⇒ AB2 = q2k2 - p2k2

⇒ AB2 = k2(q2 - p2)

⇒ AB = k2(q2−p2)=kq2−p2\sqrt{k^2(q^2 - p^2)} = k\sqrt{q^2 - p^2}.

By formula,

tan θ=PerpendicularBase=ABBC=kq2−p2pk=q2−p2p.cot θ=1tan θ=1q2−p2p=pq2−p2.\text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AB}{BC} = \dfrac{k\sqrt{q^2 - p^2}}{pk} \\[1em] = \dfrac{\sqrt{q^2 - p^2}}{p}. \\[1em] \text{cot θ} = \dfrac{1}{\text{tan θ}} \\[1em] = \dfrac{1}{\dfrac{\sqrt{q^2 - p^2}}{p}} \\[1em] = \dfrac{p}{\sqrt{q^2 - p^2}}.

Substituting values in tan θ - cot θ we get :

⇒tan θ - cot θ=q2−p2p−pq2−p2=q2−p2q2−p2−p2pq2−p2=q2−p2−p2pq2−p2=q2−2p2pq2−p2.\Rightarrow \text{tan θ - cot θ} = \dfrac{\sqrt{q^2 - p^2}}{p} - \dfrac{p}{\sqrt{q^2 - p^2}} \\[1em] = \dfrac{\sqrt{q^2 - p^2}\sqrt{q^2 - p^2} - p^2}{p\sqrt{q^2 - p^2}} \\[1em] = \dfrac{q^2 - p^2 - p^2}{p\sqrt{q^2 - p^2}} \\[1em] = \dfrac{q^2 - 2p^2}{p\sqrt{q^2 - p^2}}.

Hence, tan θ - cot θ = q2−2p2pq2−p2.\dfrac{q^2 - 2p^2}{p\sqrt{q^2 - p^2}}.

Question 4

Given 4 sin θ = 3 cos θ, find the values of :

(i) sin θ

(ii) cos θ

(iii) cot2 θ - cosec2 θ

Answer

Let ABC be a triangle with ∠B = 90° and ∠C = θ.

Given 4 sin θ = 3 cos θ, find the values of (i) sin θ (ii) cos θ (iii) cot^2 θ - cosec^2 θ. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Given,

⇒ 4 sin θ = 3 cos θ

⇒ sin θcos θ=34\dfrac{\text{sin θ}}{\text{cos θ}} = \dfrac{3}{4}.

⇒ tan θ = 34\dfrac{3}{4} ...........(1)

From figure.

⇒ tan θ = PerpendicularBase=ABBC\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{BC} ............(2)

From (1) and (2) we get :

ABBC=34\dfrac{AB}{BC} = \dfrac{3}{4}

Let AB = 3x and BC = 4x.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ AC2 = (3x)2 + (4x)2

⇒ AC2 = 9x2 + 16x2

⇒ AC2 = 25x2

⇒ AC2 = 25x2\sqrt{25x^2}

⇒ AC = 5x.

(i) By formula,

⇒ sin θ = PerpendicularHypotenuse=ABAC=3x5x=35\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{3x}{5x} = \dfrac{3}{5}.

Hence, sin θ = 35\dfrac{3}{5}.

(ii) By formula,

⇒ cos θ = BaseHypotenuse=BCAC=4x5x=45\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{4x}{5x} = \dfrac{4}{5}.

Hence, cos θ = 45\dfrac{4}{5}.

(iii) By formula,

⇒ cot θ = BasePerpendicular=BCAB=4x3x=43\dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{BC}{AB} = \dfrac{4x}{3x} = \dfrac{4}{3}.

⇒ cot2 θ = (43)2\Big(\dfrac{4}{3}\Big)^2 = 169\dfrac{16}{9}.

⇒ cosec θ = HypotenusePerpendicular=ACAB=5x3x=53\dfrac{\text{Hypotenuse}}{\text{Perpendicular}} = \dfrac{AC}{AB} = \dfrac{5x}{3x} = \dfrac{5}{3}.

⇒ cosec2 θ = (53)2\Big(\dfrac{5}{3}\Big)^2 = 259\dfrac{25}{9}.

cot2 θ−cosec2 θ=169−259=16−259=−99=−1.\text{cot}^2 \text{ θ} - \text{cosec}^2 \text{ θ} = \dfrac{16}{9} - \dfrac{25}{9} \\[1em] = \dfrac{16 - 25}{9} \\[1em] = -\dfrac{9}{9} \\[1em] = -1.

Hence, cot2 θ - cosec2 θ = -1.

Question 5

If 2 cos θ = 3\sqrt{3}, prove that 3 sin θ - 4 sin3 θ = 1.

Answer

Given,

⇒ 2 cos θ = 3\sqrt{3}

⇒ cos θ = 32\dfrac{\sqrt{3}}{2}

Squaring both sides we get :

⇒ cos2 θ = (32)2\Big(\dfrac{\sqrt{3}}{2}\Big)^2 = 34\dfrac{3}{4}.

⇒ 1 - sin2 θ = 34\dfrac{3}{4}.

⇒ sin2 θ = 1−341 - \dfrac{3}{4}.

⇒ sin2 θ = 4−34\dfrac{4 - 3}{4}.

⇒ sin2 θ = 14\dfrac{1}{4}.

⇒ sin θ = 14\sqrt{\dfrac{1}{4}}

⇒ sin θ = 12\dfrac{1}{2}.

Substituting values in L.H.S. of equation 3 sin θ - 4 sin3 θ = 1 we get :

⇒3sin θ−4 sin3θ⇒sin θ (3−4 sin2θ)⇒12×(3−4×14)⇒12×(3−1)⇒12×2⇒1.\Rightarrow 3\text{sin θ} - 4\text{ sin}^3 θ \\[1em] \Rightarrow \text{sin θ }(3 - 4\text{ sin}^2 θ) \\[1em] \Rightarrow \dfrac{1}{2} \times (3 - 4 \times \dfrac{1}{4}) \\[1em] \Rightarrow \dfrac{1}{2} \times (3 - 1) \\[1em] \Rightarrow \dfrac{1}{2} \times 2 \\[1em] \Rightarrow 1.

Since, L.H.S. = R.H.S.

Hence, proved that 3sin θ - 4 sin3 θ = 1.

Question 6

If sec θ - tan θsec θ + tan θ=14,\dfrac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} = \dfrac{1}{4}, find sin θ.

Answer

Given, sec θ - tan θsec θ + tan θ=14.\dfrac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} = \dfrac{1}{4}.

Solving above equation we get,

⇒sec θ - tan θsec θ + tan θ=14⇒1cos θ−sin θcos θ1cos θ+sin θcos θ⇒1 - sin θcos θ1 + sin θcos θ⇒1 - sin θ1 + sin θ=14⇒4(1−sin θ)=1+sin θ⇒4−4 sin θ=1+sin θ⇒sin θ + 4 sin θ=4−1⇒5 sin θ=3⇒sin θ=35.\Rightarrow \dfrac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos θ}} - \dfrac{\text{sin θ}}{\text{cos θ}}}{\dfrac{1}{\text{cos θ}} + \dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 - sin θ}}{\text{cos θ}}}{\dfrac{\text{1 + sin θ}}{{\text{cos θ}}}} \\[1em] \Rightarrow \dfrac{\text{1 - sin θ}}{\text{1 + sin θ}} = \dfrac{1}{4} \\[1em] \Rightarrow 4(1 - \text{sin θ}) = 1 + \text{sin θ} \\[1em] \Rightarrow 4 - 4\text{ sin θ} = 1 + \text{sin θ} \\[1em] \Rightarrow \text{sin θ + 4 sin θ} = 4 - 1 \\[1em] \Rightarrow 5 \text{ sin θ} = 3 \\[1em] \Rightarrow \text{sin θ} = \dfrac{3}{5}.

Hence, sin θ = 35\dfrac{3}{5}.

Question 7

If sin θ + cosec θ = 3133\dfrac{1}{3}, find the value of sin2 θ + cosec2 θ.

Answer

Given,

⇒sin θ + cosec θ=313⇒sin θ + cosec θ=103\phantom{\Rightarrow} \text{sin θ + cosec θ} = 3\dfrac{1}{3} \\[1em] \Rightarrow \text{sin θ + cosec θ} = \dfrac{10}{3} \\[1em]

Squaring both sides we get,

⇒(sin θ + cosec θ)2=(103)2⇒sin2 θ+cosec2 θ+ 2 sin θ.cosec θ=1009⇒sin2 θ+cosec2 θ+2 sin θ×1sin θ=1009⇒sin2 θ+cosec2 θ+2=1009⇒sin2 θ+cosec2 θ=1009−2⇒sin2 θ+cosec2 θ=100−189⇒sin2 θ+cosec2 θ=829⇒sin2 θ+cosec2 θ=919\Rightarrow \text{(sin θ + cosec θ)}^2 = \Big(\dfrac{10}{3}\Big)^2 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + \text{ 2 sin θ.cosec θ} = \dfrac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + \text{2 sin θ} \times \dfrac{1}{\text{sin θ}} = \dfrac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + 2 = \dfrac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \dfrac{100}{9} - 2 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \dfrac{100 - 18}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \dfrac{82}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = 9\dfrac{1}{9}

Hence, sin2 θ + cosec2 θ = 9199\dfrac{1}{9}.

Question 8

In the adjoining figure, cosec x = 135\dfrac{13}{5}, AB = 26 cm and sin y = 817\dfrac{8}{17}. Find BC.

Answer

By formula,

sin θ = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

cosec θ = HypotenusePerpendicular\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}

In ΔABD,

⇒cosec x=ABBD⇒135=26BD⇒BD=26×513⇒BD=2×5⇒BD=10 cm.\Rightarrow \text{cosec x} = \dfrac{AB}{BD}\\[1em] \Rightarrow \dfrac{13}{5} = \dfrac{26}{BD}\\[1em] \Rightarrow BD = \dfrac{26 \times 5}{13}\\[1em] \Rightarrow BD = 2 \times 5\\[1em] \Rightarrow BD = 10 \text{ cm}.

Since, ΔABD is a right angled triangle. Using pythagoras theorem,

⇒ AB2 = BD2 + AD2

⇒ 262 = 102 + AD2

⇒ 676 = 100 + AD2

⇒ AD2 = 676 - 100

⇒ AD2 = 576

⇒ AD = 576\sqrt{576}

⇒ AD = ± 24 cm

As length of side of a triangle cannot be negative. So, AD = 24 cm.

In ΔADC,

⇒sin y=ADAC⇒817=24AC⇒AC=24×178⇒AC=3×17⇒AC=51 cm.\Rightarrow \text{sin y} = \dfrac{AD}{AC}\\[1em] \Rightarrow \dfrac{8}{17} = \dfrac{24}{AC}\\[1em] \Rightarrow AC = \dfrac{24 \times 17}{8}\\[1em] \Rightarrow AC = 3 \times 17\\[1em] \Rightarrow AC = 51 \text{ cm}.

Since, ΔADC is a right angled triangle. Using pythagoras theorem,

⇒ AC2 = AD2 + DC2

⇒ 512 = 242 + DC2

⇒ 2601 = 576 + DC2

⇒ DC2 = 2601 - 576

⇒ DC2 = 2025

⇒ DC = 2025\sqrt{2025}

⇒ DC = ± 45

As length of side of a triangle cannot be negative. So, DC = 45 cm.

From figure,

BC = BD + DC = 10 + 45 = 55 cm.

Hence, the length of BC = 55 cm.

Question 9

In the adjoining figure, AB = 4 m and ED = 3 m. If sin α = 35\dfrac{3}{5} and cos β = 1213\dfrac{12}{13}, find the length of BD.

In the adjoining figure, AB = 4 m and ED = 3 m. If sin α = 3/5 and cos β = 12/13, find the length of BD. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In △ABC,

By formula,

sin α = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

Substituting values we get :

⇒35=ABAC⇒35=4AC⇒AC=203\Rightarrow \dfrac{3}{5} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{3}{5} = \dfrac{4}{AC} \\[1em] \Rightarrow AC = \dfrac{20}{3}

In right angle triangle ABC,

By pythagoras theorem, we get :

⇒ AC2 = AB2 + BC2

⇒(203)2=42+BC2⇒4009=16+BC2⇒BC2=4009−16⇒BC2=400−1449⇒BC2=2569⇒BC=25616⇒BC=163 m\Rightarrow \Big(\dfrac{20}{3}\Big)^2 = 4^2 + BC^2 \\[1em] \Rightarrow \dfrac{400}{9} = 16 + BC^2 \\[1em] \Rightarrow BC^2 = \dfrac{400}{9} - 16 \\[1em] \Rightarrow BC^2 = \dfrac{400 - 144}{9} \\[1em] \Rightarrow BC^2 = \dfrac{256}{9} \\[1em] \Rightarrow BC = \sqrt{\dfrac{256}{16}} \\[1em] \Rightarrow BC = \dfrac{16}{3} \text{ m}

In △CDE,

By formula,

cos β = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

Substituting values we get :

⇒1213=CDCE\Rightarrow \dfrac{12}{13} = \dfrac{CD}{CE}

Let CD = 12k and CE = 13k.

In right angle triangle ABC,

⇒ CE2 = CD2 + ED2

⇒ (13k)2 = (12k)2 + 32

⇒ 169k2 = 144k2 + 32

⇒ 32 = 169k2 - 144k2

⇒ 9 = 25k225k^2

⇒ k=925=35k = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}.

CD = 12k = 12×35=36512 \times \dfrac{3}{5} = \dfrac{36}{5}

From figure,

BD=BC+CD=163+365=80+10815=18815=12815 m.BD = BC + CD = \dfrac{16}{3} + \dfrac{36}{5} \\[1em] = \dfrac{80 + 108}{15} \\[1em] = \dfrac{188}{15} \\[1em] = 12\dfrac{8}{15} \text{ m}.

Hence, BD = 1281512\dfrac{8}{15} m.

PrevNext