Trigonometric Equations Question 113
Question: If in a triangle $ ABC $ , $ 2\cos A=\sin B,cosec,C, $ then
[MP PET 1996]
Options:
A) $ a=b $
B) $ b=c $
C) $ c=a $
D) $ 2a=bc $
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Answer:
Correct Answer: C
Solution:
$ 2\cos A=\frac{\sin B}{\sin C}\Rightarrow \frac{c^{2}+b^{2}-a^{2}}{bc}=\frac{b}{c} $ $ \Rightarrow $ $ c^{2}=a^{2}\Rightarrow $ $ =\pi R^{2}=4\pi R^{2} $
BETA
