Trigonometrical Ratios — Multiple Choice Questions

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.

The value of sin A is

1. $\dfrac{7}{24}$

2. $\dfrac{7}{25}$

3. $\dfrac{25}{7}$

4. $\dfrac{24}{25}$

Answer

In right angled triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (24)² + (7)²

⇒ AC² = 576 + 49

⇒ AC² = 625

⇒ AC = $\sqrt{625}$ = 25 cm.

By formula,

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

= $\dfrac{BC}{AC} = \dfrac{7}{25}$.

Hence, Option 2 is the correct option.

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.

The value of sec A is

1. $\dfrac{24}{7}$

2. $\dfrac{7}{24}$

3. $\dfrac{25}{24}$

4. $\dfrac{25}{7}$

Answer

By formula,

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

= $\dfrac{AC}{AB} = \dfrac{25}{24}$.

Hence, Option 3 is the correct option.

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.

The value of tan C is

1. $\dfrac{24}{7}$

2. $\dfrac{7}{24}$

3. $\dfrac{7}{25}$

4. $\dfrac{24}{25}$

Answer

By formula,

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

= $\dfrac{AB}{BC} = \dfrac{24}{7}$.

Hence, Option 1 is the correct option.

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.

The value of cosec C is

1. $\dfrac{7}{24}$

2. $\dfrac{24}{25}$

3. $\dfrac{25}{7}$

4. $\dfrac{25}{24}$

Answer

By formula,

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

= $\dfrac{AC}{AB} = \dfrac{25}{24}$.

Hence, Option 4 is the correct option.

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.

The value of tan A + cot C is

1. $\dfrac{7}{12}$

2. $\dfrac{12}{7}$

3. $\dfrac{14}{25}$

4. $\dfrac{25}{12}$

Answer

By formula,

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

= $\dfrac{BC}{AB} = \dfrac{7}{24}$.

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

= $\dfrac{BC}{AB} = \dfrac{7}{24}$.

Substituting value in tan A + cot C we get :

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

Hence, Option 1 is the correct option.

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.

The value of 2 cos A - sin C is

1. $\dfrac{25}{24}$

2. $\dfrac{24}{25}$

3. $\dfrac{41}{25}$

4. $\dfrac{49}{25}$

Answer

By formula,

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

= $\dfrac{AB}{AC} = \dfrac{24}{25}$.

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

= $\dfrac{AB}{AC} = \dfrac{24}{25}$.

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

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

Hence, Option 2 is the correct option.

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

1. 0

2. 1

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

4. 2

Answer

In right angle triangle ABC,

⇒ BC² = AB² + AC²

⇒ 10² = (6)² + (AC)²

⇒ AC² = 10² - 6²

⇒ AC² = 100 - 36

⇒ AC² = 64

⇒ AC = $\sqrt{64}$ = 8 cm.

By formula,

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

= $\dfrac{AC}{BC} = \dfrac{8}{10}$.

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

= $\dfrac{AB}{BC} = \dfrac{6}{10}$.

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

= $\dfrac{AB}{BC} = \dfrac{6}{10}$.

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

= $\dfrac{AC}{BC} = \dfrac{8}{10}$.

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

$\Rightarrow \text{sin B cos C + sin C cos B} = \dfrac{8}{10} \times \dfrac{8}{10} + \dfrac{6}{10} \times \dfrac{6}{10} \\[1em] = \dfrac{64}{100} + \dfrac{36}{100} \\[1em] = \dfrac{100}{100} = 1.$

Hence, Option 2 is the correct option.

In the adjoining figure, the value of cos θ is

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

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

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

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

Answer

In right angled triangle BDC,

⇒ BC² = BD² + CD²

⇒ BC² = (4)² + (3)²

⇒ BC² = 16 + 9

⇒ BC² = 25

⇒ BC = $\sqrt{25}$ = 5 cm.

In right angled triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (12)² + (5)²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = $\sqrt{169}$ = 13 cm.

By formula,

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

= $\dfrac{AB}{AC} = \dfrac{12}{13}$.

Hence, Option 1 is the correct option.

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

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

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

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

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

Answer

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

By formula,

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

Substituting values we get :

$\Rightarrow \dfrac{4}{5} = \dfrac{AB}{AC}$

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 = $\sqrt{9x^2}$ = 3x.

By formula,

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

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

Hence, Option 2 is the correct option.

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

1. $\sqrt{3}$

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

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

4. 1

Answer

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

By formula,

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

Substituting values we get :

$\Rightarrow \dfrac{1}{2} = \dfrac{BC}{AC}$

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 = $\sqrt{3x^2} = \sqrt{3}x$.

By formula,

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

Hence, Option 1 is the correct option.

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

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

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

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

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

Answer

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

By formula,

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

Substituting values we get :

$\Rightarrow \dfrac{13}{12} = \dfrac{AC}{AB}$

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 = $\sqrt{25x^2} = 5x$.

By formula,

tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{BC} = \dfrac{12x}{5x} = \dfrac{12}{5}$.

Hence, Option 1 is the correct option.

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

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

2. $\dfrac{y}{\sqrt{x^2 + y^2}}$

3. $\dfrac{x^2 - y^2}{\sqrt{x^2 + y^2}}$

4. $\dfrac{x^2 - y^2}{x^2 + y^2}$

Answer

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

By formula,

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

Substituting values we get :

$\Rightarrow \dfrac{x}{y} = \dfrac{BC}{AB}$

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 = $\sqrt{k^2(y^2 + x^2)}$

⇒ AC = $k\sqrt{y^2 + x^2}$.

By formula,

cos A = $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{yk}{k\sqrt{x^2 + y^2}} = \dfrac{y}{\sqrt{x^2 + y^2}}$.

Hence, Option 2 is the correct option.

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

1. $\dfrac{b}{\sqrt{b^2 - a^2}}$

2. $\dfrac{b}{a}$

3. $\dfrac{\sqrt{b^2 - a^2}}{b}$

4. $\dfrac{a}{\sqrt{b^2 - a^2}}$

Answer

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

By formula,

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

Substituting values we get :

$\Rightarrow \dfrac{a}{b} = \dfrac{AB}{AC}$

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 = $\sqrt{k^2(b^2 - a^2)}$

⇒ BC = $k\sqrt{(b^2 - a^2)}$.

By formula,

$\text{cos θ} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{BC}{AC} \\[1em] = \dfrac{k\sqrt{(b^2 - a^2)}}{bk} \\[1em] = \dfrac{\sqrt{b^2 - a^2}}{b}$

Hence, Option 3 is the correct option.

Consider the following two statements:

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

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 = $\dfrac{1}{2}$

By formula,

⇒ sin² A + cos² A = 1

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

⇒ $\dfrac{1}{4}$ + cos² A = 1

⇒ cos² A = $1 - \dfrac{1}{4}$

⇒ cos² A = $\dfrac{3}{4}$

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

By formula,

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

Thus, both the statements are false.

Hence, option 2 is the correct option.