question_answer

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:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

A) \[\frac{2\sin \alpha \sin \beta }{\sin \left( \beta -\alpha  \right)}\]           

B) \[\frac{\sin \alpha \sin \beta }{\cos \left( \beta -\alpha  \right)}\]

C) \[\frac{2\sin \left( \beta -\alpha  \right)}{\sin \alpha \sin \beta }\]                           

D) \[\frac{\cos \left( \beta -\alpha  \right)}{\sin \alpha \sin \beta }\]

Correct Answer: A

Solution :

                Let AB be the tower of height 'h'. Given : In\[\ln \Delta ABP\] \[\tan \alpha =\frac{AB}{PB}\]or\[\frac{\sin \alpha }{\cos \alpha }=\frac{h}{x+2}\] \[\Rightarrow \]\[(x+2)sin\alpha =h\,cos\alpha \] \[\Rightarrow \]\[h=\frac{x\sin \alpha +2\sin \alpha }{\cos \alpha }\]                                      ?(1) Now, \[\ln \Delta ABC,\tan \beta =\frac{AB}{BC}\] \[\Rightarrow \]\[\frac{\sin \beta }{\cos \beta }=\frac{h}{x}\Rightarrow x=\frac{h\cos \beta }{\sin \beta }\]                                ?(2) Putting the value of x in eq. (2) to eq. (1), we get\[h=\frac{\frac{h\cos \beta \sin \alpha }{\sin \beta }+\frac{2\sin \alpha }{1}}{\cos \alpha }\] \[\Rightarrow \]\[h=\frac{h\cos \beta .\sin \alpha +2\sin \alpha \sin \beta }{sin\beta .cos\alpha }\] \[\Rightarrow \]\[h(sin\beta .cos\alpha -cos\beta .sin\alpha )=2sin\alpha .sin\beta \] \[\Rightarrow \]\[h[sin(\beta -\alpha )]=2sin\alpha .sin\beta \] \[\Rightarrow \]\[h=\frac{2\sin \alpha .\sin \beta }{\sin (\beta -\alpha )}\]