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JEE Main & Advanced Mathematics Trigonometric Equations Question Bank Relation between sides and angles, Solutions of triangles

  • question_answer
    In a \[\Delta ABC\], \[a,\ b,\ A\]are given and \[{{c}_{1}},\ {{c}_{2}}\]are two values of the third side c. The sum of the areas of two triangles with sides \[a,\ b,\ {{c}_{1}}\] and \[a,b,\ {{c}_{2}}\] is

    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 = \[\frac{1}{2}ab\sin {{C}_{1}}+\frac{1}{2}ab\sin {{C}_{2}}\]=\[\frac{1}{2}ab(\sin {{C}_{1}}+\sin {{C}_{2}})\] \[=\frac{1}{2}ab(2\sin B\cos A)=ab\sin B\cos A\] = \[b(a\sin B)\cos A=b(b\sin A)\cos A=\frac{1}{2}{{b}^{2}}\sin 2A\].


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