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 C2bcosA cosA=b2+c2−a22bc c2−2bccosA+b2−a2=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=b2−a2