If in a triangle ABC,a=5,b=4 and cos(A−B)=3132. then the third side c is equal to
A
3
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B
6
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C
7
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D
9
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Solution
The correct option is B 6 cos(A−B)=1−tan2A−B21+tan2A−B2⇒3132=1−tan2A−B21+tan2A−B2⇒63tan2A−B2=1⇒tanA−B2=1√63 Now tanA−B2=a−ba+bcotC2⇒1√63=5−45+4cotC2⇒tanC2=√639Also cosC=1−tan2(C/2)1+tan2(C/2) =1−63811+6381=18144=18∴c2=a2+b2−2abcosC=25+16−2⋅5⋅4⋅18=36
Hence c=6.