Aptitude

Question: 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:
[A].

173 m

[B].

200 m

[C].

273 m

[D].

300 m

Answer: Option C

Explanation:

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.

Question: An observer 1.6 m tall is 203 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:
[A].

21.6 m

[B].

23.2 m

[C].

24.72 m

[D].

None of these

Answer: Option A

Explanation:

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.

Question: 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:
[A].

2.3 m

[B].

4.6 m

[C].

7.8 m

[D].

9.2 m

Answer: Option D

Explanation:

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.

Question: 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:
[A].

149 m

[B].

156 m

[C].

173 m

[D].

200 m

Answer: Option C

Explanation:

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.

Question: The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:
[A].

30°

[B].

45°

[C].

60°

[D].

90°

Answer: Option A

Explanation:

Let AB be the tree and AC be its shadow.

Let ACB = .

Then,	AC	=	3         cot = 3
	AB

= 30°.