Question

A statue 8 meters high standing on the top of a tower 64 meters high, on the bank of a river
subtends at a point A on the opposite bank, directly facing the tower, the same angle as
subtended at the same point A by man of height 2 meters standing at the base of the tower
Prove that the breadth of the river is 16V6 meters.​

Answer 1

The breadth of the river is proved to be equal to 16√6 meters.

Step-by-step explanation:

Referring to the figure attached below, let’s make some assumptions

BC = 2 m = height of the man

BD = 64 m = height of the tower

DE = 8 m = height of the statue

AB = x = breadth of the river

α = angle DAB  

β = angle EAD = angle CAB = angle subtended by the statue as well as by the man at point A

Consider ∆ABC, by applying the trigonometry ratios of a triangle, we have

tan β = perpendicular/base  

⇒ tan β = 2/x ….. (i)

Consider ∆ADB, by applying the trigonometry ratios of a triangle, we have

tan α = perpendicular/base  

⇒ tan α = 64/x ….. (ii)

Consider ∆ABE, by applying the trigonometry ratios of a triangle, we have

tan (α+β) = perpendicular/base

⇒ tan (α+β) = (64+8)/x

⇒ tan (α+β) = 72/x ….. (iii)

Now, we have the formula,

tan (α+β) = [tan α + tan β] / [1 – tanα tanβ] ….. (iv)

Substituting the values from (i), (ii) & (iii) in the formula in (iv),

$\frac{72}{x} = \frac{\frac{64}{x} + \frac{2}{x}}{1 - (\frac{64}{x}) (\frac{2}{x})}$

⇒ 72 = $\frac{64+2}{1 - (\frac{64}{x})(\frac{2}{x})}$

⇒ 72 = $\frac{66}{\frac{x^2 - 128}{x^2} }$

⇒ 72 = $\frac{66x^2}{x^2 - 128}$

⇒ 72x² – 9216 = 66x²

⇒ 6x² = 9216

⇒ x² = 9216/6

⇒ x² = 1536

⇒ x = √[6 * 16 * 16]  

⇒ x = 16√6 m

Hence it is proved that the breadth of the river is 16√6 m.

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