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In a △ABC with usual notation, if ∠A=∠B=12(sin−1(√6+12√3)+sin−1(1√3)) and c=6⋅31/4, then the area of △ABC in square units, is |
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Answer» In a △ABC with usual notation, if ∠A=∠B=12(sin−1(√6+12√3)+sin−1(1√3)) and c=6⋅31/4, then the area of △ABC in square units, is |
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