Question
In the figure given below, find RS and PS using the information given in $\Delta PSR$.
✓
Answer
In $\Delta PSR$, $\angle S = 90^{\circ}$ and $\angle P = 30^{\circ}$. Thus, $\angle R = 60^{\circ}$.
By $30^{\circ}-60^{\circ}-90^{\circ}$ triangle theorem:
Side opposite to $30^{\circ}$ is $RS = \frac{1}{2} \times PR$
$RS = \frac{1}{2} \times 12 = 6$
Side opposite to $60^{\circ}$ is $PS = \frac{\sqrt{3}}{2} \times PR$
$PS = \frac{\sqrt{3}}{2} \times 12 = 6\sqrt{3}$
By $30^{\circ}-60^{\circ}-90^{\circ}$ triangle theorem:
Side opposite to $30^{\circ}$ is $RS = \frac{1}{2} \times PR$
$RS = \frac{1}{2} \times 12 = 6$
Side opposite to $60^{\circ}$ is $PS = \frac{\sqrt{3}}{2} \times PR$
$PS = \frac{\sqrt{3}}{2} \times 12 = 6\sqrt{3}$
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