Question

Ab is a straight road leading to c, the foot of a tower. A being at the distance of 200m from c and b is 125m nearer. If the angle of elevation of the top of the tower at b is double the angle of elevation at a, find the height of the tower

Answer 1

a/c to question,
AB is straight road and C is the foot of tower.

Let DC is tower of height h.
it is given that, AC = 120m and BC = 200 - 125 = 75m as it is 125m nearer than A.

Solution : In Triangle BCD

tan2x = h/75

or, h = 75tan2x

or, h = 75{2tanx/(1 - tan²x)} ........(1)


In Triangle, ACD

tanx = h/200

h = 200tanx ...........(2)

From equations (1) and (2)

200tanx = 150tanx/(1 - tan²x)

or, 200tanx - 200tan³x = 150tanx

or, 50tanx = 200tan³x

or, 1/4 = tan²x => tanx = 1/2

hence, h = 200tanx = 200 × 1/2 = 100m

so the height of tower is 100m