question_answer

If the angle of elevation of a cloud from a point h metre above a lake is p and the angle of depression of its reflection in the lake is a, then the height of the cloud is

A)  \[\frac{h\,\,(\alpha -\beta )}{(\alpha -\beta )}\]

B)  \[h\,\,(\alpha -\beta )\sin \,\,(\alpha -\beta )\]

C)  \[h\sin \,\,(\alpha +\beta )\,\,(\alpha -\beta )\]

D)  \[\frac{h\,\,(\alpha +\beta )}{\sin \,\,(\alpha -\beta )}\]

Correct Answer: C

Solution :

Let P be the cloud and P its reflection in the lake T be the point 'h' metre above the surface of the lake. \[\therefore \]      ST = h Also,     \[NP=NP'=x\] (say)     \[PM=x-h\] \[P'M=x+h\] In \[\Delta PTM,\]\[\frac{PM}{TM}=\tan \beta \] \[\therefore \]      \[x-h=TM\tan \beta \]                  ?(i) In \[\Delta P'TM,\]\[\frac{P'M}{TM}=\tan \alpha \] \[x+h=TM\tan \alpha \]              ?(ii) From Eqs. (i) and (ii), we get \[\frac{x-h}{x+h}=\frac{\tan \beta }{\tan \alpha }\] Use componendo and dividendo rute \[\frac{x}{h}=\frac{\tan \beta +\tan \alpha }{\tan \alpha -\tan \beta }\] \[\frac{x}{h}=\frac{\sin \beta \cos \alpha +\cos \beta \sin \alpha }{\sin \alpha \cos \beta -\cos \alpha \sin \beta }=\frac{\sin \,\,(\alpha +\beta )}{\sin \,\,(\alpha -\beta )}\] \[\therefore \]\[x=h\sin \,\,(\alpha +\beta )\cdot \text{cosec}\,\,\text{(}\alpha -\beta )\]