Trigonometrical Ratios — Multiple Choice Questions

If θ is an acute angle and sin(θ - 15°) = $\dfrac{1}{2}$, then cos(θ - 15°) =

1. $\dfrac{1}{2}$

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

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

4. 1

Answer

Given,

sin(θ - 15°)= $\dfrac{1}{2}$

sin(θ - 15°) = sin 30°

θ - 15° = 30°

θ = 30° + 15° = 45°

so,

cos(θ - 15°) = cos(45° - 15°) = cos 30° = $\dfrac{\sqrt{3}}{2}$

Hence, option 2 is the correct option.

If 0° ≤θ ≤ 90° and cos(θ - 30°) = $\dfrac{1}{2}$, then tan θ =

1. $\sqrt{3}$

2. 1

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

4. undefined

Answer

Given,

cos(θ - 30°) = $\dfrac{1}{2}$

cos(θ - 30°) = cos 60°

θ - 30° = 60°

θ = 60° + 30° = 90°

Then,

tan θ = tan 90° = undefined.

Hence, option 4 is the correct option.

If x tan 30° = cos 60°, then x =

1. 2

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

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

4. $\dfrac{1}{2}$

Answer

Given,

x tan 30° = cos 60°

$x \times \dfrac{1}{\sqrt{3}} = \dfrac{1}{2}$

x = $\dfrac{\sqrt{3}}{2}$

Hence, option 3 is the correct option.

If 0° ≤θ ≤ 90° and tan (θ + 15°)= 1, then cos 2θ =

1. $\dfrac{1}{2}$

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

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

4. 0

Answer

tan (θ + 15°) = 1

tan (θ + 15°) = tan 45°

θ + 15° = 45°

θ = 45° - 15°= 30°

Then

cos 2θ = cos 2(30°) = cos 60° = $\dfrac{1}{2}$.

Hence, option 1 is the correct option.

If sin θ = cos θ, then sec (θ + 15°) =

1. $\sqrt{2}$

2. 2

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

4. 1

Answer

Given,

sin θ = cos θ

This is possible in case of θ = 45° as sin 45° = cos 45° = $\dfrac{1}{\sqrt{2}}$.

θ = 45°

Then,

sec (θ + 15°) = sec (45° + 15°)

= sec 60°

= 2.

Hence, option 2 is the correct option.

If cos 2θ = 0 and θ is an acute angle, then cot(θ - 15°) =

1. $\dfrac{1}{\sqrt{3}}$

2. 1

3. $\sqrt{3}$

4. undefined

Answer

Given,

cos 2θ = 0

cos 2θ = cos 90°

2θ = 90°

θ = 45°

Then,

cot(θ - 15°) = cot (45° - 15°) = cot 30° = $\sqrt{3}$.

Hence, option 3 is the correct option.

If θ is an acute angle and sin (θ + 18°) = $\dfrac{1}{2}$, then cosec 5θ =

1. 2

2. $\sqrt{2}$

3. 1

4. $\dfrac{2}{\sqrt{3}}$

Answer

Given,

sin(θ + 18°) = $\dfrac{1}{2}$

sin(θ + 18°) = sin 30°

θ + 18° = 30°

θ = 30° - 18° = 12°

Then,

cosec 5θ = cosec 5(12°) = cosec 60° = $\dfrac{2}{\sqrt{3}}$.

Hence, option 4 is the correct option.

If sin θ = $\dfrac{8}{17}$, then cot θ =

1. $\dfrac{15}{8}$

2. $\dfrac{15}{17}$

3. $\dfrac{8}{15}$

4. $\dfrac{17}{8}$

Answer

sin θ = $\dfrac{8}{17}$

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

Let Perpendicular = 8x and Hypotenuse = 17x

We will find Base by using Pythagoras Theorem,

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

Base² = (17x)² - (8x)²

Base² = 289x² - 64x²

Base² = 225x²

Base = 15x

Then,

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

= $\dfrac{15x}{8x} = \dfrac{15}{8}$.

Hence, option 1 is the correct option.

If sin θ = $\dfrac{1}{2}$, then (3cos θ - 4 cos³ θ)=

1. 0

2. $\dfrac{1}{2}$

3. $\dfrac{1}{6}$

4. -1

Answer

sin θ = $\dfrac{1}{2}$

sin 30° = $\dfrac{1}{2}$

sin θ = sin 30°

θ = 30°

Then,

3cos θ - 4 cos³ θ = 3 cos 30° - 4 (cos 30°)³

= $3\times \dfrac{\sqrt{3}}{2}- 4 \times \Big(\dfrac{\sqrt{3}}{2}\Big)^3$

= $\dfrac{3\sqrt{3}}{2}- \dfrac{12\sqrt{3}}{8}$

= $\dfrac{3\sqrt{3}}{2}- \dfrac{3\sqrt{3}}{2}$

= 0.

Hence, option 1 is the correct option.

If 5 cot θ = 3, then $\dfrac{(5 \sin θ - 3 \cos θ)}{(4 \sin θ + 3 \cos θ)}$ =

1. $\dfrac{11}{18}$

2. $\dfrac{16}{29}$

3. $\dfrac{14}{27}$

4. $\dfrac{29}{16}$

Answer

Given,

5 cot θ = 3

cot θ = $\dfrac{3}{5}$

cot θ = $\dfrac{\cos θ}{\sin θ}$

Given,

$\dfrac{(5 \sin θ - 3 \cos θ)}{(4 \sin θ + 3 \cos θ)}$

Dividing above equation by sin θ, we get :

$\Rightarrow \dfrac{\dfrac{5\sin θ - 3\cos θ}{\sin θ}}{\dfrac{4\sin θ + 3 \cos θ}{\sin θ}} \\[1em] \Rightarrow \dfrac{(5 - 3 \cot θ)}{(4 + 3 \cot θ)}\\[1em] = \dfrac{\Big(5 - 3\times \dfrac{3}{5}\Big)}{\Big(4 + 3\times \dfrac{3}{5}\Big)}\\[1em] = \dfrac{\Big(5 - \dfrac{9}{5}\Big)}{\Big(4 + \dfrac{9}{5}\Big)}\\[1em] = \dfrac{\Big(\dfrac{{25 -9}}{5}\Big)}{\Big( \dfrac{20 + 9}{5}\Big)}\\[1em] = \dfrac{\Big(\dfrac{{16}}{5}\Big)}{\Big( \dfrac{29}{5}\Big)}\\[1em] = \dfrac{16}{5}\times \dfrac{5}{29}\\[1em] = \dfrac{16}{29}.$

Hence, option 2 is the correct option.

In △ABC, ∠B = 90°, AB = 5 cm and BC = 12 cm. Then sin C =

1. $\dfrac{12}{13}$

2. $\dfrac{5}{13}$

3. $\dfrac{5}{12}$

4. $\dfrac{13}{5}$

Answer

Perpendicular = AB = 5 cm

Base = BC = 12 cm

By using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

Hypotenuse² = (5)² + (12)²

Hypotenuse² = 25² + 144²

Hypotenuse² = 169²

Hypotenuse = 13 cm

AC = 13 cm

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

= $\dfrac{5}{13}$.

Hence, option 2 is the correct option.

The value of sin θ cos (90° - θ) + cos θ sin(90° - θ) =

1. 0

2. 1

3. 2

4. $\dfrac{3}{2}$

Answer

Solving,

⇒ sin θ cos (90° - θ) + cos θ sin(90° - θ)

⇒ sin θ sin θ + cos θ cos θ

⇒ sin²θ + cos²θ

⇒ 1.

Hence, option 2 is the correct option.

The value of sin²25° + sin²65° =

1. 90

2. 40

3. 0

4. 1

Answer

⇒ sin²25° + sin²65°

⇒ sin²25° + sin²(90° - 25°)

⇒ sin²25° + cos²25°

⇒ 1.

Hence, option 4 is the correct option.

The value of $\dfrac{\tan 35^\circ}{\cot 55^\circ} + \dfrac{\cot 78^\circ}{\tan 12^\circ}$ =

1. 0

2. 2

3. 1

4. 3

Answer

Given,

$\Rightarrow \dfrac{\tan 35^\circ}{\cot 55^\circ} + \dfrac{\cot 78^\circ}{\tan 12^\circ}$

= $\dfrac{\tan 35^\circ}{\cot (90^\circ - 35^\circ)} + \dfrac{\cot (90^\circ - 12^\circ)}{\tan 12^\circ}$

= $\dfrac{\tan 35^\circ}{\tan 35^\circ} + \dfrac{\tan 12^\circ}{\tan 12^\circ}$

= 1 + 1 = 2.

Hence, option 2 is the correct option.

If 3 sin θ + 4 cos θ = 5, then the value of sin θ is :

1. $\dfrac{3}{4}$

2. $\dfrac{3}{5}$

3. $\dfrac{4}{5}$

4. $\dfrac{5}{3}$

Answer

3 sin θ + 4 cos θ = 5

4 cos θ = 5 - 3 sin θ

Squaring Both Sides,

(4 cos θ)² = (5 - 3 sin θ)²

16 cos²θ = 25 + 9 sin²θ - 30 sin θ

Putting cos²θ = 1 - sin²θ

16 ( 1 - sin²θ) = 25 + 9 sin²θ - 30 sin θ

16 - 16 sin²θ = 25 + 9 sin²θ - 30 sin θ

25 + 9 sin²θ - 30 sin θ - 16 + 16 sin²θ = 0

25 sin²θ - 30 sin θ + 9 = 0

25 sin²θ - 15 sin θ - 15 sin θ + 9 = 0

5 sin θ(5sin θ - 3) - 3(5sin θ - 3) = 0

(5 sin θ - 3)(5sin θ - 3)= 0

(5 sin θ - 3)² = 0

5 sin θ - 3 = 0

$\Rightarrow \sin θ = \dfrac{3}{5}$

Hence, option 2 is the correct option.

The value of tan 5° tan 25° tan 30° tan 65° tan 85° =

1. 1

2. $\sqrt{3}$

3. $\dfrac{1}{\sqrt{3}}$

4. 2

Answer

Solving,

⇒ tan 5° tan 25° tan 30° tan 65° tan 85°

⇒ tan 5° tan 85° tan 25° tan 65° tan 30°

⇒ tan 5° tan (90° - 5°) tan 25° tan (90° - 25°) tan 30°

⇒ tan 5° cot 5° tan 25° cot 25° tan 30°

⇒ $1 \times 1 \times$ tan 30°

⇒ $\dfrac{1}{\sqrt{3}}$.

Hence, option 3 is the correct option.

The value of (cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°) =

1. $\dfrac{3}{5}$

2. $\dfrac{5}{6}$

3. $\dfrac{7}{4}$

4. $\dfrac{5}{8}$

Answer

(cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°)

$\Rightarrow \Big(1 + \dfrac{1}{\sqrt{2}} + \dfrac{1}{2}\Big) \Big(1 - \dfrac{1}{\sqrt{2}} + \dfrac{1}{2}\Big)\\[1em] = \big(\dfrac{3}{2} + \dfrac{1}{\sqrt{2}}\Big) \Big(\dfrac{3}{2} - \dfrac{1}{\sqrt{2}}\Big)\\[1em] = \Big(\dfrac{3}{2}\Big)^2 - \Big(\dfrac{1}{\sqrt{2}}\Big)^2\\[1em] = \dfrac{9}{4} - \dfrac{1}{2}\\[1em] = \dfrac{9 - 2}{4} = \dfrac{7}{4}$

Hence, option 3 is the correct option.