Solution of Right Triangles — Exercise 22

The given figure, shows an isosceles triangle with angle ABC = 90°, the measure of angle BAC is :

1. 30°

2. 45°

3. 60°

4. 40°

Answer

Given that Δ ABC is isosceles, ∠ BAC = ∠ BCA.

Let ∠BAC = ∠BCA = x.

Using the angle sum property of a triangle, which states that the sum of all angles in a triangle is 180°:

⇒ ∠BAC + ∠BCA + ∠ABC = 180°

⇒ x + x + 90° = 180°

⇒ 2x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\dfrac{90°}{2}$

⇒ x = 45°

Hence, ∠BAC = 45°.

Hence, option 2 is the correct option.

In the given triangle, ∠ACB = 90° and angle CAB = 60°, the value of cot ∠ABC is :

1. 2

2. 1

3. ${\sqrt3}$

4. $\dfrac{1}{\sqrt3}$

Answer

Let ∠ABC be x.

Using the angle sum property of a triangle, which states that the sum of all angles in a triangle is 180°:

⇒ ∠CAB + ∠ ACB + ∠ ABC = 180°

⇒ 60° + 90° + x = 180°

⇒ 150° + x = 180°

⇒ x = 180° - 150°

⇒ x = 30°

Now, cot ∠ABC = cot 30°

= $\sqrt3$

Hence, option 3 is the correct option.

In the given triangle, the length of AB is :

1. $4{\sqrt2}$

2. 4 m

3. 8 m

4. 6 m

Answer

Let ∠BAC be x.

Using the angle sum property of a triangle, which states that the sum of all angles in a triangle is 180°:

⇒ ∠BAC + ∠BCA + ∠ABC = 180°

⇒ x + 45° + 90° = 180°

⇒ x + 135° = 180°

⇒ x = 180° - 135°

⇒ x = 45°

Thus, ∠BAC = ∠BCA = 45°. Therefore, Δ ABC is an isosceles triangle, meaning AB = BC = 8 m.

Hence, option 3 is the correct option.

From the information given in the triangle shown below, the length of CD is :

(take ${\sqrt3}$ = 1.732)

(i) 7.32 m

(ii) 27.32 m

(iii) 10 m

(iv) 17.32 m

Answer

In Δ ABD,

$\text{tan 45°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ 1 = \dfrac{AB}{BD}\\[1em] ⇒ 1 = \dfrac{10}{BD}\\[1em] ⇒ BD = 10 m$

In Δ ABC,

$\text{tan 30°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \dfrac{1}{\sqrt3} = \dfrac{AB}{BC}\\[1em] ⇒ \dfrac{1}{1.732} = \dfrac{10}{BC}\\[1em] ⇒ BC = 10 \times 1.732 m\\[1em] ⇒ BC = 17.32 m$

As, BC = BD + CD

⇒ 17.32 = 10 + CD

⇒ CD = 17.32 - 10

⇒ CD = 7.32

Hence, option 1 is the correct option.

In the given triangle, the length of AB is :

1. 160 cm

2. 120 cm

3. 60 cm

4. 40 cm

Answer

Let AC = x. Therefore, DC = x.

In Δ ABC,

$\text{cos 60°} = \dfrac{Base}{Hypotenuse}\\[1em] ⇒ \dfrac{1}{2} = \dfrac{BC}{x}\\[1em] ⇒ BC = \dfrac{x}{2}$

$\text{sin 60°} = \dfrac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \dfrac{\sqrt3}{2} = \dfrac{AB}{x}\\[1em] ⇒ AB = \dfrac{x\sqrt3}{2}$

$BD = BC + CD\\[1em] ⇒ BD = \dfrac{x}{2} + x\\[1em] ⇒ BD = \dfrac{x + 2x}{2}\\[1em] ⇒ BD = \dfrac{3x}{2}$

In Δ ABD, according to Pythagoras theorem,

⇒ AD² = BD² + AB² (∵ AD is hypotenuse)

$⇒ 80^2 = \Big(\dfrac{3x}{2}\Big)^2 + \Big(\dfrac{\sqrt3x}{2}\Big)^2 \\[1em] ⇒ 6400 = \dfrac{9x^2}{4} + \dfrac{3x^2}{4} \\[1em] ⇒ 6400 = \dfrac{9x^2 + 3x^2}{4}\\[1em] ⇒ 6400 = \dfrac{12x^2}{4}\\[1em] ⇒ 6400 = 3x^2\\[1em] ⇒ x^2 = \dfrac{6400}{3}\\[1em] ⇒ x = \sqrt\dfrac{6400}{3}\\[1em] ⇒ x = \dfrac{80}{\sqrt3}$

Substituting the value of x in AB,

$AB = \dfrac{x\sqrt3}{2}\\[1em] ⇒ AB = \dfrac{\dfrac{80}{\sqrt3} \times \sqrt3}{2}\\[1em] ⇒ AB = \dfrac{\dfrac{80}{\cancel {\sqrt3}} \times \cancel {\sqrt3}}{2}\\[1em] ⇒ AB = 40 m$

Hence, option 4 is the correct option.

Find 'x', if :

Answer

$\text{sin 60°} = \dfrac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \dfrac{\sqrt3}{2} = \dfrac{20}{x}\\[1em] ⇒ x = \dfrac{20 \times 2}{\sqrt3}\\[1em] ⇒ x = \dfrac{40}{\sqrt3} \\[1em] ⇒ x = 23.1$

Hence, x = 23.1.

Find 'x', if :

Answer

$\text{tan 30°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \dfrac{1}{\sqrt3} = \dfrac{20}{x}\\[1em] ⇒ x = 20 \times \sqrt3\\[1em] ⇒ x = 34.64$

Hence, x = 34.64.

Find 'x', if :

Answer

$\text{sin 45°} = \dfrac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \dfrac{1}{\sqrt2} = \dfrac{20}{x}\\[1em] ⇒ x = 20 \times \sqrt2\\[1em] ⇒ x = 28.28$

Hence, x = 28.28.

Find angle 'A' if :

Answer

$\text{cos A} = \dfrac{Base}{Hypotenuse}\\[1em] ⇒ \text{cos A} = \dfrac{10}{20}\\[1em] ⇒ \text{cos A} = \dfrac{1}{2}\\[1em] ⇒ \text{cos A} = \text{cos 60°}$

Hence, A = 60°.

Find angle 'A' if :

Answer

$\text{sin A} = \dfrac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \text{sin A} = \dfrac{\dfrac{10}{\sqrt2}}{10}\\[1em] ⇒ \text{sin A} = \dfrac{1}{\sqrt2}\\[1em] ⇒ \text{sin A} = \text{sin 45°}$

Hence, A = 45°.

Find angle 'A' if :

Answer

$\text{tan A} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \text{tan A} = \dfrac{10\sqrt3}{10}\\[1em] ⇒ \text{tan A} = \sqrt3\\[1em] ⇒ \text{tan A} = \text{tan 60°}$

Hence, A = 60°.

Find angle 'x' if :

Answer

In Δ ABC,

$\text{tan 60°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \dfrac{CB}{AC}\\[1em] ⇒ \sqrt3 = \dfrac{30}{AC}\\[1em] ⇒ AC = \dfrac{30}{\sqrt3}\\[1em] ⇒ AC = \dfrac{30 \times \sqrt3}{3}\\[1em] ⇒ AC = 10\sqrt3$

In Δ ADC,

$\text{sin x} = \dfrac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \text{sin x} = \dfrac{AC}{AD}\\[1em] ⇒ \text{sin x} = \dfrac{10\sqrt3}{20}\\[1em] ⇒ \text{sin x} = \dfrac{\sqrt3}{2}\\[1em] ⇒ \text{sin x} = \text {sin 60°}$

Hence, x = 60°.

Find AD, if :

Answer

BE = CD = 50 m

ED = BC = 10 m

In Δ ABE,

$\text{tan 45°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ 1 = \dfrac{AE}{BE}\\[1em] ⇒ 1 = \dfrac{AE}{50}\\[1em] ⇒ AE = 50$

AD = AE + ED

⇒ AD = 50 + 10 m

⇒ AD = 60 m

Hence, AD = 60 m.

Find AD, if :

Answer

Let AC = x. Therefore, BC = x.

In Δ ADC,

$\text{cos 60°} = \dfrac{Base}{Hypotenuse}\\[1em] ⇒ \dfrac{1}{2} = \dfrac{CD}{x}\\[1em] ⇒ CD = \dfrac{x}{2}\\[1em]$

And,

$\text{sin 60°} = \dfrac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \dfrac{\sqrt3}{2} = \dfrac{AD}{x}\\[1em] ⇒ AD = \dfrac{x\sqrt3}{2}\\[1em]$

$\text{BD = BC + CD}\\[1em] ⇒ BD = x + \dfrac{x}{2}\\[1em] ⇒ BD = \dfrac{2x + x}{2}\\[1em] ⇒ BD = \dfrac{3x}{2}$

In Δ ABD, according to Pythagoras theorem,

⇒ AB² = BD² + AD² (∵ AB is hypotenuse)

$⇒ 100^2 = \Big(\dfrac{3x}{2}\Big)^2 + \Big(\dfrac{\sqrt3x}{2}\Big)^2 \\[1em] ⇒ 10000 = \dfrac{9x^2}{4} + \dfrac{3x^2}{4} \\[1em] ⇒ 10000 = \dfrac{9x^2 + 3x^2}{4}\\[1em] ⇒ 10000 = \dfrac{12x^2}{4}\\[1em] ⇒ 10000 = 3x^2\\[1em] ⇒ x^2 = \dfrac{10000}{3}\\[1em] ⇒ x = \sqrt\dfrac{10000}{3}\\[1em] ⇒ x = \dfrac{100}{\sqrt3}$

Substituting the value of x in AD,

$AD = \dfrac{x\sqrt3}{2}\\[1em] ⇒ AD = \dfrac{\dfrac{100}{\sqrt3} \times \sqrt3}{2}\\[1em] ⇒ AD = \dfrac{\dfrac{100}{\cancel {\sqrt3}} \times \cancel {\sqrt3}}{2}\\[1em] ⇒ AD = 50 m$

Hence, AD = 50 m.

Find the length of AD.

Given :

∠ABC = 60°,
∠DBC = 45°
and BC = 40 cm.

Answer

In Δ BDC,

$\text{tan 45°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ 1 = \dfrac{DC}{BC}\\[1em] ⇒ 1 = \dfrac{DC}{40}\\[1em] ⇒ DC = 40$

In Δ ABC,

$\text{tan 60°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \dfrac{AC}{BC}\\[1em] ⇒ \sqrt3 = \dfrac{AC}{40}\\[1em] ⇒ AC = 40\sqrt3 = 69.28$

AD = AC - DC

⇒ AD = 69.28 - 40 = 29.28 cm

Hence, AD = 29.28 cm.

Find lengths of diagonals AC and BD.

Given AB = 60 cm and ∠BAD = 60°.

Answer

ABCD is a rhombus as AB = BC = CD = DA and ∠ BAD = 60°.

Now in Δ ABD,

AB = AD (Sides of rhombus)

⇒ ∠ ABD = ∠ ADB (Angles opposite to equal side of Δ)

Let ∠ ABD = x.

According to the angle sum property,

∠ ABD + ∠ ADB + ∠ BAD = 180°

⇒ x + x + 60° = 180°

⇒ 2x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$

⇒ x = 60°

ABCD is a rhombus. So, diagonals AC and BD bisect each other at 90°.

AO = OC and BO = OD

In Δ AOD,

$\text{cos 60°} = \dfrac{Base}{Hypotenuse}\\[1em] ⇒ \dfrac{1}{2} = \dfrac{OD}{60}\\[1em] ⇒ OD = \dfrac{60}{2} = 30$

And,

$\text{sin 60°} = \dfrac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \dfrac{\sqrt3}{2} = \dfrac{AO}{60}\\[1em] ⇒ AO = \dfrac{60\sqrt3}{2} = 51.96$

AC = 2 x AO = 2 x 51.96 = 103.92 cm

BD = 2 x BO = 2 x 30 = 60 cm

Hence, AC = 103.92 cm and BD = 60 cm.

Find AB.

Answer

Draw a line FP parallel to CD such that FP = CD and FC = PD.

In Δ ACF,

$\text{tan 45°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ 1 = \dfrac{20}{AC}\\[1em] ⇒ AC = 20$

In Δ DEB,

$\text{tan 60°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \dfrac{30}{BD}\\[1em] ⇒ BD = \dfrac{30}{\sqrt3}\\[1em] ⇒ BD = \dfrac{30 \times \sqrt3}{\sqrt3 \times \sqrt3}\\[1em] ⇒ BD = \dfrac{30\sqrt3}{3}\\[1em] ⇒ BD = 10\sqrt3 = 17.32$

As, FC = 20 and ED = 30

EP = ED - PD = ED - FC = 30 - 20 = 10 cm

In Δ EFP,

$\text{tan 60°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \dfrac{FP}{10}\\[1em] ⇒ FP = 10 \sqrt3 = 17.32$

Thus AB = AC + CD + BD

= 20 + 17.32 + 17.32

= 54.64

Hence, AB = 54.64.

In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and ∠A = 60°.

Find :

(i) length of AB

(ii) distance between AB and DC.

Answer

(i) Draw two perpendicular lines, DP and CM, from points D and C to AB, respectively. Since AB is parallel to CD, PMCD will form a rectangle.

In Δ APD,

cos 60° = $\dfrac{Base}{Hypotenuse}$

⇒ $\dfrac{1}{2} = \dfrac{AP}{AD}$

⇒ $\dfrac{1}{2} = \dfrac{AP}{20}$

⇒ AP = 10

Now, Δ APD and Δ BMC,

DP = CM (Height of same quadrilateral)

AD = CB (Both are 20 cm)

∠ DPA = ∠ CMB

So, Δ APD ≅ Δ BMC (∵ RHS congruency)

And, by using corresponding sides of congruent triangle,

AP = MB = 10 cm

PM = DC = 20 cm

AB = AP + PM + MB

= 10 + 20 + 10 cm

= 40 cm

Hence, AB = 40 cm.

(ii) In Δ APD,

sin 60° = $\dfrac{Perpendicular}{Hypotenuse}$

⇒ $\dfrac{\sqrt3}{2} = \dfrac{PD}{AD}$

⇒ $\dfrac{\sqrt3}{2} = \dfrac{PD}{20}$

⇒ PD = 10 $\sqrt3$ = 17.32 cm

Hence, PD = 17.32 cm.

Use the information given to find the length of AB.

Answer

In Δ APQ,

$\text{tan 30°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \dfrac{1}{\sqrt3} = \dfrac{AQ}{AP}\\[1em] ⇒ \dfrac{1}{\sqrt3} = \dfrac{10}{AP}\\[1em] ⇒ AP = 10 \sqrt3 = 17.32 cm$

In Δ PBR,

$\text{tan 45°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ 1 = \dfrac{PB}{BR}\\[1em] ⇒ 1 = \dfrac{PB}{8}\\[1em] ⇒ PB = 8 cm$

AB = AP + PB

= 17.32 + 8 cm

= 25.32 cm

Hence, AB = 25.32 cm.