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Question
In \[∆ ABC, if a = 5, b = 6 a\text{ and } C = 60°\] show that its area is \[\frac{15\sqrt{3}}{2} sq\].units.
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Solution
\[Given: a = 5, b = 6, c = 60°\]
\[\text{ Area of a triangle } = \frac{1}{2}ab\sin C\]
\[ = \frac{1}{2} \times 5 \times 6 \times \sin60°= 15 \times \frac{\sqrt{3}}{2}sq . \text{ units }\]
\[\text{ Hence proved } .\]
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