Question:
In $\Delta A B C$, if $a=\sqrt{2}, b=\sqrt{3}$ and $c=\sqrt{5}$, show that its area is $\frac{1}{2} \sqrt{6}$ sq. units.
Solution:
Given: $a=\sqrt{2}, b=\sqrt{3}, c=\sqrt{5}$
$\because \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$
$\Rightarrow \cos C=\frac{2+3-5}{2 \times \sqrt{6}}=0$
$\Rightarrow \cos C=0$
$\Rightarrow \cos C=\cos 90^{\circ}$
$\Rightarrow C=90^{\circ}$
Thus, $\sin C=\sin 90^{\circ}=1$
Hence, Area of $\Delta A B C=\frac{1}{2} a b \sin C=\frac{1}{2} \sqrt{6} \times 1=\frac{\sqrt{6}}{2}$ sq. units.
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