Competitive Exams : Quantitative Aptitude - Trig (Set 2) | Concept of Maths

Question 1

1. A $10\text{ m}$ long ladder is placed against a vertical wall such that it reaches a window $5\text{ m}$ high from the ground. What is the angle of inclination of the ladder with the horizontal ground?

A) $60^\circ$

B) $30^\circ$

C) $45^\circ$

D) $90^\circ$

Answer

Correct Answer: B

Let the angle with the ground be $\theta$. In the right triangle formed, the ladder is the hypotenuse ($H = 10\text{ m}$) and the window height is the opposite side ($P = 5\text{ m}$). $$\sin \theta = \frac{P}{H} = \frac{5}{10} = \frac{1}{2}$$ Since $\sin 30^\circ = \frac{1}{2}$, we have $\theta = 30^\circ$.

Question 2

2. If $\tan(2\theta) = \cot(\theta + 30^\circ)$ where $2\theta$ and $(\theta + 30^\circ)$ are acute angles, find the value of $\theta$.

A) $15^\circ$

B) $25^\circ$

C) $20^\circ$

D) $30^\circ$

Answer

Correct Answer: C

We know that $\tan A = \cot B$ if $A + B = 90^\circ$. So, $2\theta + (\theta + 30^\circ) = 90^\circ$ $3\theta + 30^\circ = 90^\circ \implies 3\theta = 60^\circ$ $\theta = 20^\circ$.

Question 3

3. What is the value of the expression $(\sec^2 \theta - 1) \cdot \cot^2 \theta$?

A) $1$

B) $0$

C) $\tan^2 \theta$

D) $\cos^2 \theta$

Answer

Correct Answer: A

Using the identity $1 + \tan^2 \theta = \sec^2 \theta$, we know $\sec^2 \theta - 1 = \tan^2 \theta$. The expression becomes: $\tan^2 \theta \cdot \cot^2 \theta$ Since $\cot \theta = \frac{1}{\tan \theta}$, we have $\tan^2 \theta \cdot \frac{1}{\tan^2 \theta} = 1$.

Question 4

4. A vertical pole $6\text{ m}$ high casts a shadow of length $2\sqrt{3}\text{ m}$ on the ground. Find the angle of elevation of the Sun.

A) $30^\circ$

B) $45^\circ$

C) $60^\circ$

D) $90^\circ$

Answer

Correct Answer: C

Let the angle of elevation be $\theta$. $\tan \theta = \frac{\text{Height of Pole}}{\text{Length of Shadow}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$. Since $\tan 60^\circ = \sqrt{3}$, the angle of elevation is $60^\circ$.

Question 5

5. If $5\sin \theta = 3$, find the value of $\tan \theta$ (assume $\theta$ is in the first quadrant).

A) $3/5$

B) $4/3$

C) $5/4$

D) $3/4$

Answer

Correct Answer: D

Given $\sin \theta = 3/5$, we have Perpendicular ($P$) = $3$ and Hypotenuse ($H$) = $5$. By Pythagoras theorem, Base ($B$) = $\sqrt{H^2 - P^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = 4$. $\tan \theta = \frac{P}{B} = \frac{3}{4}$.

Question 6

6. Find the value of the product: $\cos 1^\circ \cdot \cos 2^\circ \cdot \cos 3^\circ \cdot \dots \cdot \cos 100^\circ$.

A) $1$

B) $0$

C) $1/2$

D) $-1$

Answer

Correct Answer: B

The series includes the term $\cos 90^\circ$. Since $\cos 90^\circ = 0$, the entire product becomes zero regardless of the other values. $\text{Product} = \cos 1^\circ \cdot \dots \cdot 0 \cdot \dots \cdot \cos 100^\circ = 0$.

Question 7

7. From a point on the ground $30\text{ m}$ away from the foot of a tower, the angle of elevation of the top of the tower is $30^\circ$. What is the height of the tower?

A) $10\sqrt{3}\text{ m}$

B) $30\sqrt{3}\text{ m}$

C) $20\text{ m}$

D) $15\text{ m}$

Answer

Correct Answer: A

Let $h$ be the height of the tower. In the right triangle: $\tan 30^\circ = \frac{\text{Height}}{\text{Distance}} = \frac{h}{30}$ $\frac{1}{\sqrt{3}} = \frac{h}{30} \implies h = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3}\text{ m}$.

Question 8

8. Which of the following is equivalent to $\sin^2 45^\circ + \cos^2 45^\circ$?

A) $1/2$

B) $\sqrt{2}$

C) $1$

D) $0$

Answer

Correct Answer: C

By the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ for any angle $\theta$. Alternatively, $\sin 45^\circ = 1/\sqrt{2}$ and $\cos 45^\circ = 1/\sqrt{2}$. $(1/\sqrt{2})^2 + (1/\sqrt{2})^2 = 1/2 + 1/2 = 1$.

Question 9

9. If $\sin \alpha \cdot \csc \alpha + \cos \alpha \cdot \sec \alpha = x$, find $x$.

A) $1$

B) $2$

C) $0$

D) $\tan \alpha$

Answer

Correct Answer: B

Since $\csc \alpha = \frac{1}{\sin \alpha}$ and $\sec \alpha = \frac{1}{\cos \alpha}$: $\sin \alpha \cdot \frac{1}{\sin \alpha} + \cos \alpha \cdot \frac{1}{\cos \alpha} = 1 + 1 = 2$.

Question 10

10. In $\Delta ABC$ right-angled at $B$, if $\tan A = 1$, find the value of $2\sin A \cos A$.

A) $1/2$

B) $2$

C) $\sqrt{2}$

D) $1$

Answer

Correct Answer: D

If $\tan A = 1$, then $A = 45^\circ$. So, $2\sin A \cos A = 2 \sin 45^\circ \cos 45^\circ$ $= 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 2 \cdot \frac{1}{2} = 1$.

Competitive Exams : Quantitative Aptitude - Trig (Set 2) | Concept of Maths