A) \[{{b}_{1}}+{{b}_{2}}=2c\cos A\]
B) \[{{b}_{1}}+{{b}_{2}}=c\cos A\]
C) \[{{b}_{1}}+{{b}_{2}}=3c\cos A\]
D) \[{{b}_{1}}+{{b}_{2}}=4c\sin A\]
Correct Answer: A
Solution :
From cosine formula, \[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] or \[{{b}^{2}}-(2c\cos A)b+({{c}^{2}}-{{a}^{2}})=0,\] which is quadratic equation in b.\[\because c\sin A<a<c\] \[\therefore \] Two triangles will be obtained, but this is possible when two values of third side are also obtained. Clearly two values of side b will be b1 and \[{{b}_{2}}\]. Let these are roots of above equation. \[\therefore \] Sum of roots \[={{b}_{1}}+{{b}_{2}}\]= \[2c\cos A\].
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