Trigonometrical Ratios — Chapter Test

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

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 = $\sqrt{5}$.

For angle A,

$\Rightarrow \text{sin A} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{BC}{AC} = \dfrac{2}{3}. \\[1em] \Rightarrow \text{cos A} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{AB}{AC} = \dfrac{\sqrt{5}}{3}. \\[1em] \Rightarrow \text{tan A} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{BC}{AB} = \dfrac{2}{\sqrt{5}}. \\[1em] \Rightarrow \text{cot A} = \dfrac{\text{Base}}{\text{Perpendicular}} \\[1em] = \dfrac{AB}{BC} = \dfrac{\sqrt{5}}{2}. \\[1em] \Rightarrow \text{sec A} = \dfrac{\text{Hypotenuse}}{\text{Base}} \\[1em] = \dfrac{AC}{AB} = \dfrac{3}{\sqrt{5}}. \\[1em] \Rightarrow \text{cosec A} = \dfrac{\text{Hypotenuse}}{\text{Perpendicular}} \\[1em] = \dfrac{AC}{BC} = \dfrac{3}{2}. \\[1em]$

For angle C,

$\Rightarrow \text{sin C} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{AB}{AC} = \dfrac{\sqrt{5}}{3}. \\[1em] \Rightarrow \text{cos C} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{BC}{AC} = \dfrac{2}{3}. \\[1em] \Rightarrow \text{tan C} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AB}{BC} = \dfrac{\sqrt{5}}{2}. \\[1em] \Rightarrow \text{cot C} = \dfrac{\text{Base}}{\text{Perpendicular}} \\[1em] = \dfrac{BC}{AB} = \dfrac{2}{\sqrt{5}}. \\[1em] \Rightarrow \text{sec C} = \dfrac{\text{Hypotenuse}}{\text{Base}} \\[1em] = \dfrac{AC}{BC} = \dfrac{3}{2}. \\[1em] \Rightarrow \text{cosec C} = \dfrac{\text{Hypotenuse}}{\text{Perpendicular}} \\[1em] = \dfrac{AC}{AB} = \dfrac{3}{\sqrt{5}}. \\[1em]$

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

Answer

From figure,

$\Rightarrow \text{cot θ} = \dfrac{\text{Base}}{\text{Perpendicular}} \\[1em] \Rightarrow \text{cot θ} = \dfrac{AB}{BC} \\[1em] \Rightarrow \text{cot θ} = \dfrac{x}{10} \\[1em] \Rightarrow x = 10 \text{ cot θ}. \\[1em] \Rightarrow \text{cosec θ} = \dfrac{\text{Hypotenuse}}{\text{Perpendicular}} \\[1em] \Rightarrow \text{cosec θ} = \dfrac{AC}{BC} \\[1em] \Rightarrow \text{cosec θ} = \dfrac{y}{10} \\[1em] \Rightarrow y = 10 \text{ cosec θ}.$

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

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

(i) sin ∠ABC

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

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 = $\sqrt{256}$ = 16.

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

Hence, sin ∠ABC = $\dfrac{4}{5}$.

(ii) In right angle triangle BCD,

By pythagoras theorem we get :

⇒ BD² = BC² + CD²

⇒ BD² = 12² + 9²

⇒ BD² = 144 + 81

⇒ BD² = 225

⇒ BD = $\sqrt{225}$

⇒ BD = 15.

By formula,

$\text{tan x } = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{BC}{CD} = \dfrac{12}{9} = \dfrac{4}{3}. \\[1em] \text{cos x } = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{CD}{BD} = \dfrac{9}{15} = \dfrac{3}{5}. \\[1em] \text{sin x } = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{BC}{BD} = \dfrac{12}{15} = \dfrac{4}{5}. \\[1em]$

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

$\Rightarrow \dfrac{4}{3} - \dfrac{3}{5} + 3 \times \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{4}{3} - \dfrac{3}{5} + \dfrac{12}{5} \\[1em] \Rightarrow \dfrac{20 - 9 + 36}{15} \\[1em] \Rightarrow \dfrac{47}{15} \\[1em] \Rightarrow 3\dfrac{2}{15}.$

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

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

Answer

(i) By formula,

$\text{sin x } = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{AD}{AB} = \dfrac{15}{25} = \dfrac{3}{5}. \\[1em] 5\text{ sin x} = 5 \times \dfrac{3}{5} = 3.$

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 = $\sqrt{400}$ = 20.

By formula,

$\text{tan x } = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AD}{BD} = \dfrac{15}{20} = \dfrac{3}{4}. \\[1em] 7\text{ tan x} = 7 \times \dfrac{3}{4} = \dfrac{21}{4} = 5\dfrac{1}{4}.$

Hence, 7 tan x = $5\dfrac{1}{4}$.

(iii) In right angle triangle ADC,

⇒ AC² = AD² + DC²

⇒ 17² = 15² + DC²

⇒ 289 = 225 + DC²

⇒ DC² = 289 - 225

⇒ DC² = 64

⇒ DC = $\sqrt{64}$ = 8.

Solving,

$\text{5 cos x - 17 sin y - tan x} = 5 \times \dfrac{BD}{AB} - 17 \times \dfrac{CD}{AC} - \dfrac{AD}{BD} \\[1em] = 5 \times \dfrac{20}{25} - 17 \times \dfrac{8}{17} - \dfrac{15}{20} \\[1em] = 4 - 8 - \dfrac{3}{4} \\[1em] = -4 - \dfrac{3}{4} \\[1em] = \dfrac{-16 - 3}{4}\\[1em] = -\dfrac{19}{4} \\[1em] = -4\dfrac{3}{4}.$

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

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.

Given,

⇒ q cos θ = p

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

By formula,

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

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

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

$\Rightarrow \dfrac{p}{q} = \dfrac{BC}{AC}$

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 = $\sqrt{k^2(q^2 - p^2)} = k\sqrt{q^2 - p^2}$.

By formula,

$\text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AB}{BC} = \dfrac{k\sqrt{q^2 - p^2}}{pk} \\[1em] = \dfrac{\sqrt{q^2 - p^2}}{p}. \\[1em] \text{cot θ} = \dfrac{1}{\text{tan θ}} \\[1em] = \dfrac{1}{\dfrac{\sqrt{q^2 - p^2}}{p}} \\[1em] = \dfrac{p}{\sqrt{q^2 - p^2}}.$

Substituting values in tan θ - cot θ we get :

$\Rightarrow \text{tan θ - cot θ} = \dfrac{\sqrt{q^2 - p^2}}{p} - \dfrac{p}{\sqrt{q^2 - p^2}} \\[1em] = \dfrac{\sqrt{q^2 - p^2}\sqrt{q^2 - p^2} - p^2}{p\sqrt{q^2 - p^2}} \\[1em] = \dfrac{q^2 - p^2 - p^2}{p\sqrt{q^2 - p^2}} \\[1em] = \dfrac{q^2 - 2p^2}{p\sqrt{q^2 - p^2}}.$

Hence, tan θ - cot θ = $\dfrac{q^2 - 2p^2}{p\sqrt{q^2 - p^2}}.$

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

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

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

From figure.

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

From (1) and (2) we get :

$\dfrac{AB}{BC} = \dfrac{3}{4}$

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 = $\sqrt{25x^2}$

⇒ AC = 5x.

(i) By formula,

⇒ sin θ = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{3x}{5x} = \dfrac{3}{5}$.

Hence, sin θ = $\dfrac{3}{5}$.

(ii) By formula,

⇒ cos θ = $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{4x}{5x} = \dfrac{4}{5}$.

Hence, cos θ = $\dfrac{4}{5}$.

(iii) By formula,

⇒ cot θ = $\dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{BC}{AB} = \dfrac{4x}{3x} = \dfrac{4}{3}$.

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

⇒ cosec θ = $\dfrac{\text{Hypotenuse}}{\text{Perpendicular}} = \dfrac{AC}{AB} = \dfrac{5x}{3x} = \dfrac{5}{3}$.

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

$\text{cot}^2 \text{ θ} - \text{cosec}^2 \text{ θ} = \dfrac{16}{9} - \dfrac{25}{9} \\[1em] = \dfrac{16 - 25}{9} \\[1em] = -\dfrac{9}{9} \\[1em] = -1.$

Hence, cot² θ - cosec² θ = -1.

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

Answer

Given,

⇒ 2 cos θ = $\sqrt{3}$

⇒ cos θ = $\dfrac{\sqrt{3}}{2}$

Squaring both sides we get :

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

⇒ 1 - sin² θ = $\dfrac{3}{4}$.

⇒ sin² θ = $1 - \dfrac{3}{4}$.

⇒ sin² θ = $\dfrac{4 - 3}{4}$.

⇒ sin² θ = $\dfrac{1}{4}$.

⇒ sin θ = $\sqrt{\dfrac{1}{4}}$

⇒ sin θ = $\dfrac{1}{2}$.

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

$\Rightarrow 3\text{sin θ} - 4\text{ sin}^3 θ \\[1em] \Rightarrow \text{sin θ }(3 - 4\text{ sin}^2 θ) \\[1em] \Rightarrow \dfrac{1}{2} \times (3 - 4 \times \dfrac{1}{4}) \\[1em] \Rightarrow \dfrac{1}{2} \times (3 - 1) \\[1em] \Rightarrow \dfrac{1}{2} \times 2 \\[1em] \Rightarrow 1.$

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

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

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

Answer

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

Solving above equation we get,

$\Rightarrow \dfrac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos θ}} - \dfrac{\text{sin θ}}{\text{cos θ}}}{\dfrac{1}{\text{cos θ}} + \dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 - sin θ}}{\text{cos θ}}}{\dfrac{\text{1 + sin θ}}{{\text{cos θ}}}} \\[1em] \Rightarrow \dfrac{\text{1 - sin θ}}{\text{1 + sin θ}} = \dfrac{1}{4} \\[1em] \Rightarrow 4(1 - \text{sin θ}) = 1 + \text{sin θ} \\[1em] \Rightarrow 4 - 4\text{ sin θ} = 1 + \text{sin θ} \\[1em] \Rightarrow \text{sin θ + 4 sin θ} = 4 - 1 \\[1em] \Rightarrow 5 \text{ sin θ} = 3 \\[1em] \Rightarrow \text{sin θ} = \dfrac{3}{5}.$

Hence, sin θ = $\dfrac{3}{5}$.

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

Answer

Given,

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

Squaring both sides we get,

$\Rightarrow \text{(sin θ + cosec θ)}^2 = \Big(\dfrac{10}{3}\Big)^2 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + \text{ 2 sin θ.cosec θ} = \dfrac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + \text{2 sin θ} \times \dfrac{1}{\text{sin θ}} = \dfrac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + 2 = \dfrac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \dfrac{100}{9} - 2 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \dfrac{100 - 18}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \dfrac{82}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = 9\dfrac{1}{9}$

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

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

Answer

By formula,

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

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

In ΔABD,

$\Rightarrow \text{cosec x} = \dfrac{AB}{BD}\\[1em] \Rightarrow \dfrac{13}{5} = \dfrac{26}{BD}\\[1em] \Rightarrow BD = \dfrac{26 \times 5}{13}\\[1em] \Rightarrow BD = 2 \times 5\\[1em] \Rightarrow BD = 10 \text{ cm}.$

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 = $\sqrt{576}$

⇒ AD = ± 24 cm

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

In ΔADC,

$\Rightarrow \text{sin y} = \dfrac{AD}{AC}\\[1em] \Rightarrow \dfrac{8}{17} = \dfrac{24}{AC}\\[1em] \Rightarrow AC = \dfrac{24 \times 17}{8}\\[1em] \Rightarrow AC = 3 \times 17\\[1em] \Rightarrow AC = 51 \text{ cm}.$

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 = $\sqrt{2025}$

⇒ 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.

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

Answer

In △ABC,

By formula,

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

Substituting values we get :

$\Rightarrow \dfrac{3}{5} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{3}{5} = \dfrac{4}{AC} \\[1em] \Rightarrow AC = \dfrac{20}{3}$

In right angle triangle ABC,

By pythagoras theorem, we get :

⇒ AC² = AB² + BC²

$\Rightarrow \Big(\dfrac{20}{3}\Big)^2 = 4^2 + BC^2 \\[1em] \Rightarrow \dfrac{400}{9} = 16 + BC^2 \\[1em] \Rightarrow BC^2 = \dfrac{400}{9} - 16 \\[1em] \Rightarrow BC^2 = \dfrac{400 - 144}{9} \\[1em] \Rightarrow BC^2 = \dfrac{256}{9} \\[1em] \Rightarrow BC = \sqrt{\dfrac{256}{16}} \\[1em] \Rightarrow BC = \dfrac{16}{3} \text{ m}$

In △CDE,

By formula,

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

Substituting values we get :

$\Rightarrow \dfrac{12}{13} = \dfrac{CD}{CE}$

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 = $25k^2$

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

CD = 12k = $12 \times \dfrac{3}{5} = \dfrac{36}{5}$

From figure,

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

Hence, BD = $12\dfrac{8}{15}$ m.