Trigonometric Equations Question 249
Question: In a triangle $ ABC $ , $ \frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}= $ $ \frac{a}{bc}+\frac{b}{ca} $ , then the value of angle A is
[IIT 1993]
Options:
A) $ 45^{o} $
B) $ 30^{o} $
C) $ 90^{o} $
D) $ 60^{o} $
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Answer:
Correct Answer: C
Solution:
- $ \frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ca} $
Þ $ \frac{2(b^{2}+c^{2}-a^{2})}{2abc}+\frac{a^{2}+c^{2}-b^{2}}{2abc}+\frac{2(a^{2}+b^{2}-c^{2})}{2abc} $ $ =\frac{a}{bc}+\frac{b}{ca} $
Þ $ \frac{3b^{2}+c^{2}+a^{2}}{2abc} $ $ =\frac{a}{bc}+\frac{b}{ca} $ Þ $ \frac{3b}{2ac}+\frac{c}{2ab}+\frac{a}{2bc}=\frac{a}{bc}+\frac{b}{ca} $
Þ $ b^{2}+c^{2}=a^{2} $ . Hence $ \angle A=90^{o} $ .
BETA
