A) \[\frac{1}{2}{{b}^{2}}\sin 2A\]
B) \[\frac{1}{2}{{a}^{2}}\sin 2A\]
C) \[{{b}^{2}}\sin 2A\]
D) None of these
Correct Answer: A
Solution :
We have \[\cos A=\frac{{{c}^{2}}+{{b}^{2}}-{{a}^{2}}}{2bc}\] \[\Rightarrow {{c}^{2}}-2bc\,\cos A+({{b}^{2}}-{{a}^{2}})=0\] It is given that \[{{c}_{1}}\] and \[{{c}_{2}}\] are roots of this equation. Therefore \[{{c}_{1}}+{{c}_{2}}=2b\cos A\] and \[{{c}_{1}}{{c}_{2}}={{b}^{2}}-{{a}^{2}}\] \[\Rightarrow \] \[k\,(\sin {{C}_{1}}+\sin {{C}_{2}})=2k\sin B\cos A\] \[\Rightarrow \] \[\sin {{C}_{1}}+\sin {{C}_{2}}=2\sin B\cos A\] \[\Rightarrow \] Now sum of the areas of two triangles = \[\frac{1}{2}ab\sin {{C}_{1}}+\frac{1}{2}ab\sin {{C}_{2}}\]=\[\frac{1}{2}ab(\sin {{C}_{1}}+\sin {{C}_{2}})\] \[=\frac{1}{2}ab(2\sin B\cos A)=ab\sin B\cos A\] = \[b(a\sin B)\cos A=b(b\sin A)\cos A=\frac{1}{2}{{b}^{2}}\sin 2A\].You need to login to perform this action.
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