A) \[a\text{ }tan\text{ }\alpha \]
B) \[a\text{ }sin\text{ }\alpha \]
C) \[a\text{ }sin\text{ 2}\alpha \]
D) \[a\text{ }sin\text{ 3}\alpha \]
Correct Answer: C
Solution :
In \[\Delta ABE,\] \[\angle BAE=\angle AEB\] \[\therefore \] \[AB=BE\] In \[\Delta \,DCE,\] \[\sin \,3\alpha =\frac{h}{CE}\] \[\Rightarrow \] \[\sin \,3\alpha =\frac{h}{\alpha \,\sin \,2\alpha /\sin \,3\alpha }\] [from Eq. (i)] \[\Rightarrow \] \[h=a\,\sin \,2\alpha \]You need to login to perform this action.
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