IBPS PO Quantitative Aptitude: Trigonometry questions with solutions

Q1. Two ships are sailing in the sea on opposite sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships to be 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:

1. 173 m
2. 200 m
3. 273 m
4. 300 m

Answer: 273 m

For the ship at 30°, distance from the lighthouse base is 100/tan 30° = 100√3 m. For the ship at 45°, distance is 100/tan 45° = 100 m. Since the ships are on opposite sides, total distance = 100√3 + 100 ≈ 273 m.

Q2. A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30° with the man's eye. The man walks some distance towards the tower to watch its top and the angle of elevation becomes 60°. What is the distance between the base of the tower and the point P?

1. 43 units
2. 8 units
3. 12 units
4. Data inadequate

Answer: Data inadequate

Let the tower height be h and the initial distance from P be x. From the first position, tan 30° = h/x, and from the second position, tan 60° = h/(x-d), where d is the distance walked. Since both h and d are unknown, x cannot be uniquely determined. Therefore, the data is inadequate.

Q3. The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

1. 2.3 m
2. 4.6 m
3. 7.8 m
4. 9.2 m

Answer: 9.2 m

The foot of the ladder is 4.6 m from the wall, which is the adjacent side to the 60° angle. Using cos 60° = adjacent/hypotenuse = 4.6/L, we get L = 4.6/0.5 = 9.2 m.

Q4. An observer 1.6 m tall is 20 m away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is:

1. 21.6 m
2. 23.2 m
3. 24.72 m
4. None of these

Answer: 21.6 m

The horizontal distance is 20 m and the angle of elevation is 30°. So the height above the observer’s eye is 20 tan 30° = 20/√3 ≈ 11.55 m. Adding the observer’s height 1.6 m gives about 13.15 m, which does not match the options; the source answer key indicates 21.6 m, suggesting the distance may be OCR-corrupted.

Q5. From a point P on level ground, the angle of elevation of the top of a tower is $30^\circ$. If the tower is 100 m high, the distance of point P from the foot of the tower is:

1. 149 m
2. 156 m
3. 173 m
4. 200 m

Answer: 173 m

In the right triangle, $\tan 30^\circ = \frac{\text{height}}{\text{distance}} = \frac{100}{d}$. Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we get $d = 100\sqrt{3} \approx 173$ m.