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Question

If two sides a,b and angle A be such that two triangles are formed, then the sum of two values of the third side is

A
2bsinA
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B
2bcosA
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C
b/acosA
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D
(c+b)cosA
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Solution

The correct option is C 2bcosA
cosA=b2+c2a22bc
c22bccosA+b2a2=0...(let c1 ,c2 be the two roots of this equation then using the quadratic equation propert) we can say that,
c1+c2=2bcosA
c1c2=b2a2

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