Find side $c$ of the triangle $A B C$ , given that $A = 30^{0}, B = 60^{0}$ and $b = 20 \sqrt{3}$

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Solution

$C = 180^{0} - (A + B) = 90^{0}$
$b = A C = 20 \sqrt{3}$
Now $\frac{A C}{A B} = sin 60^{0}$
$∴ \frac{20 \sqrt{3}}{A B} = \frac{\sqrt{3}}{2}$ or $\sqrt{3} A B = 40 \sqrt{3}$
$∴ A B = 40 = c$

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