Trigonometric Equations Question 98
Question: In a $ \Delta ABC $ , $ \frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c} $ and the side $ a=2, $ then area of the triangle is
[IIT Screening 1993; MP PET 2000]
Options:
A) 1
B) 2
C) $ \frac{\sqrt{3}}{2} $
D) $ \sqrt{3} $
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Answer:
Correct Answer: D
Solution:
- $ \frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}\Rightarrow \frac{\cos A}{k\sin A}=\frac{\cos B}{k\sin B}=\frac{\cos C}{k\sin C} $
Þ $ \cot A=\cot B=\cot C\Rightarrow A=B=C=60^{o} $
Þ $ \Delta ABC $ is equilateral. \ $ \Delta =\frac{\sqrt{3}}{4}a^{2}=\sqrt{3} $ .
BETA
