Trigonometric Equations Question 57
Question: In a $ \Delta ABC $ $ a,\ c,A $ are given and $ b_1,\ b_2 $ are two values of the third side b such that $ b_2=2b_1 $ . Then $ \sin A= $
Options:
A) $ \sqrt{\frac{9a^{2}-c^{2}}{8a^{2}}} $
B) $ \sqrt{\frac{9a^{2}-c^{2}}{8c^{2}}} $
C) $ \sqrt{\frac{9a^{2}+c^{2}}{8a^{2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- We have $ \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc} $
$ \Rightarrow $ $ b^{2}-2bc\cos A+(c^{2}-a^{2})=0 $ It is given that $ b_1 $ and $ b_2 $ are roots of this equation. Therefore $ b_1+b_2 $ = $ 2c\cos A $ and $ b_1b_2=c^{2}-a^{2} $
$ \Rightarrow $ $ 3b_1=2c\cos A $ , $ 2b_1^{2}=c^{2}-a^{2} $ , $ (\because b_2=2b_1 $ given)
$ \Rightarrow $ $ 2,{{( \frac{2c}{3}\cos A )}^{2}}=c^{2}-a^{2}\Rightarrow 8c^{2}(1-{{\sin }^{2}}A)=9c^{2}-9a^{2} $
$ \Rightarrow $ $ \sin A=\sqrt{\frac{9a^{2}-c^{2}}{8c^{2}}} $ .
BETA
