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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 - Mathematics

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प्रश्न

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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उत्तर

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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Sine and Cosine Formulae and Their Applications
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 10: Sine and cosine formulae and their applications - Exercise 10.1 [पृष्ठ १४]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 30 | पृष्ठ १४

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