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 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   
ANS: C - 273 m

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.

2).
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?
A   43 √units   
B   8 units   
C   12 units   
D   Data inadequate   
E   None of these   
ANS: D - Data inadequate

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

3).
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   
ANS: D - 9.2 m

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.

4).
An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The heights of the tower is
A   21.6 m   
B   23.2 m   
C   24.72 m   
D   None of these   
ANS: A - 21.6 m

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.
Let AB be the observer and CD be the tower.

Draw BE  CD.
Then, CE = AB = 1.6 m,
      BE = AC = 20√3 m.

DE = tan 30° = 1
BE √3

 DE =   20√3   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 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   
ANS: C - 173 m

Let AB be the tower.

Then, APB = 30° and AB = 100 m.

AB = tan 30° = 1
AP √3

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

6).
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°   
ANS: A - 30°

Let AB be the tree and AC be its shadow.

Let ACB = .

Then, AC = √3          cot  = √3
AB

  = 30°.