# Height and distance

1.

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 30º and 45º 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

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100 m, ACB = 30º and ADB = 45º.

AB     = tan 30º =     1             AC = AB x 3 = 1003 m.
AC     3

AB     = tan 45º = 1         AD = AB = 100 m.

CD = (AC + AD)     = (1003 + 100) m
= 100(3 + 1)
= (100 x 2.73) m
= 273 m.

2.

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man’s eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?
A.     43 units    B.     8 units
C.     12 units    D.     Data inadequate
E.     None of these

Explanation:

One of AB, AD and CD must have given.

3.

The angle of elevation of a ladder leaning against a wall is 60º 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

Explanation:

Let AB be the wall and BC be the ladder.

Then, ACB = 60º and AC = 4.6 m.

AC     = cos 60º =     1
BC     2

BC     = 2 x AC
= (2 x 4.6) m
= 9.2 m.

4.

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 30º. The heights of the tower is:
A.     21.6 m    B.     23.2 m
C.     24.72 m    D.     None of these

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 30º =     1
BE     3

DE =     203     m = 20 m.
3

CD = CE + DE = (1.6 + 20) m = 21.6 m.

5.

From a point P on a level ground, the angle of elevation of the top tower is 30º. 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

Explanation:

Let AB be the tower.

Then, APB = 30º and AB = 100 m.

AB     = tan 30º =     1
AP     3

AP     = (AB x 3) m
= 1003 m
= (100 x 1.73) m
= 173 m.