Question:
In a triangle $A B C$, if $a=2, b=4$ and $A+B=\frac{2 \pi}{3}$, then area of $\triangle A B C$ is ______________________
Solution:
Since $A+B=\frac{2 \pi}{3}$ given
and $A+B+C=\pi \quad$ (Angle sum property)
$\Rightarrow C=\pi-\frac{2 \pi}{3}$
$C=\frac{\pi}{3}$
Since area of triangle $\mathrm{ABC}=\frac{1}{2} a b \sin C$
$=\frac{1}{2} \times 2 \times 4 \times \sin \frac{\pi}{3} \quad(\therefore a=2$ and $b=4$ given $)$
$=4 \times \frac{\sqrt{3}}{2}$
$=3 \sqrt{3}$ sq. units
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