Question 1

1. A 10 m long ladder is placed against a vertical wall such that it reaches a window 5 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: Option B

Explanation: Let the angle with the ground be $\theta$. In the right triangle formed, the ladder is the hypotenuse (H = 10m) and the window height is the opposite side (P = 5m). $$\sin \theta = \frac{P}{H} = \frac{5}{10} = \frac{1}{2}$$ 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) $15^\circ$ $25^\circ$

Question 3

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

A) 1B) 0C) $\tan^2 \theta$D) $\cos^2 \theta$

Answer

✓ Correct Answer: Option A

Explanation: 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 m high casts a shadow of length $2\sqrt{3}m$ 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: Option C

Explanation: Let the angle of elevation be $\theta$. $\tan \theta = \frac{Height of Pole}{Length of Shadow} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$. 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) 3/5B) 4/3C) 5/4D) 3/4

Answer

✓ Correct Answer: Option D

Explanation: 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) 1B) 0C) 1/2D) -1

Answer

✓ Correct Answer: Option B

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

Question 7

7. From a point on the ground 30 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) $10\sqrt{3}m$B) $30\sqrt{3}m$C) 20 mD) 15 m

Answer

✓ Correct Answer: Option A

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

Question 8

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

A) 1/2B) $\sqrt{2}$C) 1D) 0

Answer

✓ Correct Answer: Option C

Explanation: 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) 1B) 2C) 0D) $\tan \alpha$

Answer

✓ Correct Answer: Option B

Explanation: 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) 1/2B) 2C) $\sqrt{2}$D) 1

Answer

✓ Correct Answer: Option D

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

Accepted Answer

✓ Correct Answer: Option C

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

Competitive Exams : Quantitative Aptitude - Trig (Set 2)

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

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

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

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

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

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

7. From a point on the ground 30 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?

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

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

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