Question

From a window ( $h$ metres high above the ground) of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are $θ_{1}$ and $θ_{2}$ respectively. Show that the height of the opposite house is $h (1 + \frac{\cot θ_{2}}{\cot θ_{1}})$ .

Accepted Answer

Given,

Let $W$ be the point on the window where the angles of elevation and depression are measured.

Let $P Q$ be the house on the opposite side.

Then $W A$ is the width of the street.

Height of the window $= h$ metres $= A Q [W R = A Q]$

Let $P A = x$ metres.

In right triangle $P A W, \tan θ_{1} = \frac{A P}{A W}$
$\Rightarrow \tan θ_{1} = \frac{x}{A W}$

So, $A W = \frac{x}{\tan θ_{1}}$

we get, $A W = x \cot θ_{1} . . . . . . . . . . . (1)$

In right triangle $Q A W, \tan θ_{2} = \frac{A Q}{A W}$

$\Rightarrow \tan θ_{2} = \frac{h}{A W}$

we get, $A W = h \cot θ_{2} . . . . . . . . . . . . (2)$

From $(1)$ and $(2)$ we get, $x \cot θ_{1} = h \cot θ_{2}$

$\Rightarrow x = h \frac{\cot θ_{2}}{\cot θ_{1}}$

Therefore, height of the opposite house $= P A + A Q$

$= x + h$

$= h \frac{\cot θ_{2}}{\cot θ_{1}} + h$

$= h (1 + \frac{\cot θ_{2}}{\cot θ_{1}})$

So, the height of the opposite house is $h (1 + \frac{\cot θ_{2}}{\cot θ_{1}})$ .

Hence, it is proved.

Tamil Nadu Board Solutions for Chapter: Trigonometry, Exercise 33: Solved Example

Tamil Nadu Board Mathematics Solutions for Exercise - Tamil Nadu Board Solutions for Chapter: Trigonometry, Exercise 33: Solved Example

Questions from Tamil Nadu Board Solutions for Chapter: Trigonometry, Exercise 33: Solved Example with Hints & Solutions

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