Question

AB is a straight road leading to C. The foot of the tower A being at a distance from C and D 125 nearer. If the angle of elevation of the tower at B be the double the angle of elevation at A. Find the height of tower.

Answer 1

Answer:

Step-by-step explanation:

Let DC be the tower of height h.

It is given that, AC = 120m and BC = 200 - 125 = 75m as it is 125m nearer than A.

In Triangle BCD

tan(2x) = $\frac{h}{75}$

h = 75tan(2x)

In Triangle, ACD

tan(x) = $\frac{h}{200}$

h = 200tanx

From equations the two equations above, we have:

200tan(x) = $\frac{150tg(x)}{1 - tan^{2} x}$

hence, h = 200tan(x) = 200 × 1/2 = 100m

so the height of tower is 100m