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 \[\frac{x}{h}=\frac{\tan \beta +\tan \alpha }{\tan \alpha -\tan \beta }\] \[\frac{x}{h}=\frac{\sin \beta \cos \alpha +\cos \beta \sin \alpha }{\sin \alpha \cos \beta -\cos \alpha \sin \beta }=\frac{\sin \,\,(\alpha +\beta )}{\sin \,\,(\alpha -\beta )}\] \[\therefore \]\[x=h\sin \,\,(\alpha +\beta )\cdot \text{cosec}\,\,\text{(}\alpha -\beta )\]You need to login to perform this action.
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