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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.
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