The angle of elevation of the top of a vertical tower from a point P on the horizontal ground was observed to be \alpha. After moving a distance 2 metres from P towards the foot of the tower, the angle of elevation changes to \beta. Then the height (in metres) of the tower is :

  • Option 1)

    \frac{2\, sin\, \alpha sin\, \beta }{sin(\beta -\alpha )}

  • Option 2)

    \frac{\, sin\, \alpha sin\, \beta }{cos(\beta -\alpha )}

  • Option 3)

    \frac{2\, sin(\beta -\alpha ) }{ sin\, \alpha\, sin\, \beta}

  • Option 4)

    \frac{cos(\beta -\alpha )}{\, sin\, \alpha \, sin\, \beta}

 

Answers (1)

As we learnt in 

Height and Distances -

The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.

-

 In \bigtriangleup APB, \: tan\alpha =\frac{AB}{BP}

  \Rightarrow BP=\frac{h}{\tan \alpha }=\frac{h\cos \alpha }{\sin \alpha }

In \bigtriangleup AQB, \: tan\beta =\frac{AB}{BQ}

 \Rightarrow BQ=\frac{h\cos \beta }{\sin \beta }

Now, PQ=2

\Rightarrow PB-QB=2

\Rightarrow h\left [ \frac{\cos \alpha }{\sin \alpha } -\frac{\cos \beta }{\sin \beta }\right ]=2

\Rightarrow h\frac{\left [ \sin (\beta -\alpha ) \right ]}{\sin \alpha \: \sin \beta }=2

\Rightarrow h=\frac{2\sin \alpha \sin \beta }{\sin \left ( \beta -\alpha \right )}


Option 1)

\frac{2\, sin\, \alpha sin\, \beta }{sin(\beta -\alpha )}

This option is correct

Option 2)

\frac{\, sin\, \alpha sin\, \beta }{cos(\beta -\alpha )}

This option is incorrect

Option 3)

\frac{2\, sin(\beta -\alpha ) }{ sin\, \alpha\, sin\, \beta}

This option is incorrect

Option 4)

\frac{cos(\beta -\alpha )}{\, sin\, \alpha \, sin\, \beta}

This option is incorrect

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