In a △ABC,$$a=6$$,$$b=3$$ and $$cos$$⁡$$C=32$$, then the area of triangle $$=$$

Given $$cos$$⁡$$C=32$$⇒$$sin$$⁡$$C=12$$ Area of triangle $$=12absin$$⁡$$C=12$$×$$6$$×$$3$$×$$12=92$$

Properties of triangles

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$In a △ ABC, a = 6, b = 3 and \cos C = \frac{\sqrt{3}}{2}, then the area of triangle =$

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Solution

$\begin{array}{l} Given \cos C = \frac{\sqrt{3}}{2} \Rightarrow \sin C = \frac{1}{2} \\ Area of triangle = \frac{1}{2} absin C \\ = \frac{1}{2} \times 6 \times 3 \times \frac{1}{2} = \frac{9}{2} \end{array}$

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