At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is 5/12. On walking 192 metres towards the tower, the tangent of the angle of elevation is 3/4. The height of the tower is

According to the question,

Let 'h' be the height of the vertical tower. And

$$\angle$$ADC = P, $$\angle$$ACB = Q, CB = $$x$$ 

In triangle ABC,

tan Q = $$\frac{AB}{BC}$$

$$\frac{3}{4} = \frac{h}{x}$$

$$x = \frac{4h}{3}$$................(1)

In triangle ABD,

tan P = $$\frac{h}{192 + x}$$

$$\frac{5}{12} = \frac{h}{192 + (4h/3)}$$

$$\frac{5}{12} = \frac{3h}{576 + 4h}$$

$$5(576 + 4h) = 36h$$

$$16h = 2880$$ (or) $$h = 180 $$m

Hence, option B is the correct answer.