A) 2 sin A
B) 2 cos A
C) \[\frac{1}{2}\cos A\]
D) \[\tan A\]
Correct Answer: B
Solution :
\[A=r+0\] and \[\mu =\frac{\sin i}{\sin r}\] \[\Rightarrow \mu =\frac{\sin 2A}{\sin A}\] \[=\frac{2\sin A\cos A}{\sin A}=2\cos A\]You need to login to perform this action.
You will be redirected in
3 sec