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Question

In a ABC with usual notation, if A=B=12(sin1(6+123)+sin1(13)) and c=631/4, then the area of ABC in square units, is

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Solution

Let sinθ=6+123


2A=2B=sin1(6+123)+sin1(13)
2A=2B=tan1(1+632)+tan1(12)
=cot1(321+32)+tan112
=π2[tan13tan12]+tan112
=π2tan13+tan12+cot12
=πtan13=2π3
2A=2B=2π3
A=B=π3
C=π3
ABC is equilateral.
Hence, area =34×c2=34×36×3=27 sq. units

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