## height and distance ###### 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.

###### 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

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

###### 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.

###### 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.

###### 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.

###### 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°. 