A) \[3\Delta \]
B) \[2\Delta \]
C) \[4\Delta \]
D) \[-4\Delta \]
Correct Answer: C
Solution :
\[\Delta =\frac{1}{2}bc\sin A\Rightarrow \frac{1}{2}{{k}^{2}}\sin B\sin C\sin A=\Delta \] \[{{a}^{2}}\sin 2B+{{b}^{2}}\sin 2A=2({{a}^{2}}\sin B\cos B+{{b}^{2}}\sin A\cos A)\] \[=2{{k}^{2}}({{\sin }^{2}}A\sin B\cos B+{{\sin }^{2}}B\sin A\cos A)\] \[=2{{k}^{2}}(\sin A\sin B)(\sin C)\] \[=2{{k}^{2}}(\sin A\sin B\sin C)=4\Delta \].
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