Question
In $\triangle\text{PQR},$ right-angled at Q, PQ = 3cm and PR = 6cm. Determine $\angle\text{P}$ and $\angle\text{R}.$
✓
Answer
From above figure$\sin\text{R}=\frac{\text{PQ}}{\text{PR}}$
$\sin\text{R}=\frac{3}{6}=\frac{1}{2}$
$\therefore\ \sin\text{R}=\sin30^\circ$
$\text{R}=\sin30^\circ$
We Know in $\triangle\text{le }\angle\text{P}+\angle\text{Q}+\angle\text{R}=180^\circ$
$\angle\text{P}+90^\circ+30^\circ=180^\circ$
$\angle\text{P}=60^\circ$
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