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Question
A person observes the angle of elevation of the peak of a hill from a station to be α. He walks c metres along a slope inclined at an angle β and finds the angle of elevation of the peak of the hill to be ϒ. Show that the height of the peak above the ground is \[\frac{c \sin \alpha \sin \left( \gamma - \beta \right)}{\left( \sin \gamma - \alpha \right)}\]
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Solution
Suppose, AB is a peak whose height above the ground is t+x.
\[In \bigtriangleup DFC, \]
\[\sin\beta = \frac{x}{c} \]
\[ \Rightarrow x = c\sin\beta\]
\[and \]
\[\tan\beta = \frac{x}{y}\]
\[ \Rightarrow y = \frac{x}{\tan\beta} = \frac{c\sin\beta}{\sin\beta} \times \cos\beta = c\cos\beta . . . \left( 1 \right)\]
\[In ∆ ADE, \]
\[\tan\gamma = \frac{t}{z}\]
\[ \Rightarrow z = t \cot\gamma . . . \left( 2 \right)\]
\[\]
\[In ∆ ABC, \]
\[\tan\alpha = \frac{t + x}{y + z}\]
\[\]
\[ \Rightarrow t + x = \left( c\cos\beta\tan\alpha + tcot\gamma \right)\tan\alpha \left( from \left( 1 \right) and \left( 2 \right) \right)\]
\[ \Rightarrow t - tcot\gamma\tan\alpha = c\cos\beta\tan\alpha - c\sin\beta \left( \because x = c\sin\beta \right)\]
\[ \Rightarrow t\left( 1 - \frac{\sin\alpha\cos\gamma}{\cos\alpha\sin\gamma} \right) = c\left( \frac{\cos\beta\sin\alpha - \cos\alpha\sin\beta}{\cos\alpha} \right)\]
\[ \Rightarrow t\left( \frac{\sin\gamma\cos\alpha - \sin\alpha\cos\gamma}{\cos\alpha\sin\gamma} \right) = c\frac{\sin\left( \alpha - \beta \right)}{\cos\alpha}\]
\[ \Rightarrow t\frac{\sin\left( \gamma - \beta \right)}{\cos\alpha\sin\gamma} = c\frac{\sin\left( \alpha - \beta \right)}{\cos\alpha}\]
\[ \Rightarrow t = c\frac{\sin\gamma\sin\left( \alpha - \beta \right)}{\sin\left( \gamma - \beta \right)} . . . \left( 3 \right)\]
\[\text{ Now }, \]
\[AB = t + x = c\frac{\sin\gamma\sin\left( \alpha - \beta \right)}{\sin\left( \gamma - \beta \right)} + c\sin\beta \left( \text{ using }\left( 3 \right) \right)\]
\[ = c\left( \frac{\sin\gamma\sin\left( \alpha - \beta \right)}{\sin\left( \gamma - \beta \right)} + \sin\beta \right)\]
\[ = c\left[ \frac{\sin\gamma\sin\left( \alpha - \beta \right) + \sin\beta\sin\left( \gamma - \beta \right)}{\sin\left( \gamma - \beta \right)} \right]\]
\[ = c\left[ \frac{\sin\gamma\sin\alpha\cos\beta - \sin\beta\sin\gamma\cos\alpha + \sin\beta\sin\gamma\cos\alpha - \sin\beta\cos\gamma\sin\alpha}{\sin\left( \gamma - \beta \right)} \right]\]
\[ = c\left[ \frac{\sin\gamma\sin\alpha\cos\beta - \sin\beta\cos\gamma\sin\alpha}{\sin\left( \gamma - \beta \right)} \right]\]
\[ = c\left[ \frac{\sin\alpha\sin\left( \gamma - \beta \right)}{\sin\left( \gamma - \beta \right)} \right]\]
\[ = \frac{c\sin\alpha\sin\left( \gamma - \beta \right)}{\sin\left( \gamma - \beta \right)}\]
\[\text{ Hence proved } . \]
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