A) \[r\,\text{cosec}\frac{\alpha }{2}\sin \beta \]
B) \[\frac{r\,\text{cosec}\,\alpha }{2\sin \beta }\]
C) \[\frac{r\,\text{cosec}\,\alpha }{2\cos \beta }\]
D) \[r\sin \alpha \sin \beta \]
Correct Answer: A
Solution :
If P is the eye, then \[\angle CPQ=\beta ,\,\angle CPA=\angle CPB=\frac{\alpha }{2}\] Let CR = h be the height of centre of balloon From right angled \[\Delta \,CAP,\] \[\sin \frac{\alpha }{2}=\frac{CA}{CP}=\frac{r}{CP}\] \[\therefore \] \[CP=r\,\text{cosec}\,\frac{\alpha }{2}\] Also, from right angled \[\Delta \,CPR,\] \[\sin \beta =\frac{h}{CP}=\frac{h}{r\,\text{cosec}\frac{\alpha }{2}}\] \[\therefore \] \[h=r\,\,\text{cosec}\frac{\alpha }{2}\sin \beta \]You need to login to perform this action.
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