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In $\triangle$ $ABC,$ right-angled at $B$, $AB =5\, cm$ and $\angle ACB =30^{\circ}$ (see $Fig.$). Determine the lengths of the sides $BC$ and $AC .$
Solution
To find the length of the side $BC ,$ we will choose the trigonometric ratio involving $BC$ and the given side $AB$. since $BC$ is the side adjacent to angle $C$ and $AB$ is the side opposite to angle $C ,$ therefore
$\frac{ AB }{ BC }=\tan C$
$\frac{5}{ BC }=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$
which gives $BC =5 \sqrt{3} \,cm$
To find the length of the side $AC ,$ we consider
$\sin 30^{\circ}=\frac{ AB }{ AC }$
$\frac{1}{2}=\frac{5}{ AC }$
$AC =10 \,cm$
Note that alternatively we could have used Pythagoras theorem to determine the third side in the example above,
$AC =\sqrt{ AB ^{2}+ BC ^{2}}=\sqrt{5^{2}+(5 \sqrt{3})^{2}} cm =10 \,cm$
