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B
12
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C
11
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D
10
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Solution
The correct option is A 9 We have, tan(A−B2)=(a−ba+b)cotC2 ⇒√1−cos(A−B)1+cos(A−B)=(a−ba+b)cotC2 ⇒√1−(4/5)1+(4/5)=(6−36+3)cotC2 ⇒cotC2=1 ⇒C=π2 now, △=absinC2=12×6×3×sinπ2=9 Ans: A