Trigonometric Equations Question 316
Question: If the area of a triangle ABC is D, then $ a^{2}\sin 2B+b^{2}\sin 2A $ is equal to
[WB JEE 1988]
Options:
A) $ 3\Delta $
B) $ 2\Delta $
C) $ 4\Delta $
D) $ -4\Delta $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \Delta =\frac{1}{2}bc\sin A\Rightarrow \frac{1}{2}k^{2}\sin B\sin C\sin A=\Delta $ $ a^{2}\sin 2B+b^{2}\sin 2A=2(a^{2}\sin B\cos B+b^{2}\sin A\cos A) $ $ =2k^{2}({{\sin }^{2}}A\sin B\cos B+{{\sin }^{2}}B\sin A\cos A) $ $ =2k^{2}(\sin A\sin B)(\sin C) $ $ =2k^{2}(\sin A\sin B\sin C)=4\Delta $ .
BETA
