Triangles And Properties Of Triangle Question 8
Question: If in a $ \Delta ABC,\cos AsinB=sinC $ then the value of $ \tan \frac{A}{2}, $ if $ 3b-5c=0 $ , is
Options:
A) 0.5
B) 0.75
C) 0.33
D) $ \frac{1}{\sqrt{3}} $
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Answer:
Correct Answer: A
Solution:
[a] $ \cos A\sin B=\sin C\Rightarrow \sin (A+B)-sin(A-B)=2sinC $
$ \Rightarrow \sin C=\sin (B-A)\Rightarrow A+C=B $ $ (\because ,A+B=\pi -C) $
$ \therefore ,B=\frac{\pi }{2} $ Now $ 3b-5c=0\Rightarrow 3-5\sin C=0 $
$ \therefore \sin C=\frac{3}{5}andA=\frac{\pi }{2}-C $
$ \Rightarrow \cos A=\sin C\Rightarrow \frac{1-{{\tan }^{2}}\frac{A}{2}}{1+{{\tan }^{2}}\frac{A}{2}}=\frac{3}{5} $
$ \therefore {{\tan }^{2}}\frac{A}{2}=\frac{1}{4}\Rightarrow \tan \frac{A}{2}=0.5 $
BETA
