Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

An observer 1.6 m tall is 20 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:

Answer» Let AB be the observer and CD be the tower. Draw BE CD. Then, CE = AB = 1.6 m,       BE = AC = 203 m. DE = tan 30° = 1 BE 3 DE = 203 m = 20 m. 3 CD = CE + DE = (1.6 + 20) m = 21.6 m.
Discussion
2.

The angle of elevation of the sun, when the length of the shadow of a tree times the height of the tree, is:

Answer» Let AB be the tree and AC be its shadow. Let ACB = . Then, AC = 3         cot = 3 AB = 30°.
Discussion
3.

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

Answer» Let AB be the lighthouse and C and D be the positions of the ships. Then, AB = 100 m, ACB = 30° and ADB = 45°. AB = tan 30° = 1         AC = AB x 3 = 1003 m. AC 3 AB = tan 45° = 1         AD = AB = 100 m. AD CD = (AC + AD) = (1003 + 100) m = 100(3 + 1) = (100 x 2.73) m = 273 m.
Discussion
4.

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 the elevation becomes 60°. What is the distance between the base of the tower and the point P?

Answer» One of AB, AD and CD must have given. So, the data is inadequate.
Discussion
5.

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:

Answer» Let AB be the wall and BC be the ladder. Then, ACB = 60° and AC = 4.6 m. AC = cos 60° = 1 BC 2 BC = 2 x AC = (2 x 4.6) m = 9.2 m.
Discussion
6.

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

Answer» Let AB be the tower. Then, APB = 30° and AB = 100 m. AB = tan 30° = 1 AP 3 AP = (AB x 3) m = 1003 m = (100 x 1.73) m = 173 m.
Discussion