In $ \Delta P Q R $, right-angled at $ Q, P Q=3 \mathrm{~cm} $ and $ P R=6 \mathrm{~cm} $. Determine $ \angle P $ and $ \angle R $.
Given:
In \( \Delta P Q R \), right-angled at \( Q, P Q=3 \mathrm{~cm} \) and \( P R=6 \mathrm{~cm} \).
To do:
We have to determine \( \angle P \) and \( \angle R \).
Solution:
In \( \Delta P Q R \),
$sin\ R=\frac{PQ}{PR}$
$=\frac{3}{6}$
$=\frac{1}{2}$
$=\sin 30^{\circ}$ (Since $\sin 30^{\circ}=\frac{1}{2}$)
$\Rightarrow R=30^{\circ}$
We know that sum of the angles in a triangle is $180^o$.
Therefore,
$\angle P+\angle Q+\angle R=180^o$
$\angle P+90^o+30^o=180^o$
$\angle P=180^o-120^o$
$\angle P=60^o$
Hence, $\angle P=60^o$ and $\angle R=30^o$.
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