In triangle PQR,right-angled at Q,PQ = 3 , cm | vedclass

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Question

(A) Given that in $	riangle PQR$,$\angle Q = 90^{\circ}$,$PQ = 3 \, cm$,and $PR = 6 \, cm$.For $\angle PRQ$ (or $\angle R$),the side opposite is $PQ$

Vedclass · Accepted Answer

$\angle QPR = 60^{\circ}, \angle PRQ = 30^{\circ}$

Vedclass · Answer

(A) Given that in $	riangle PQR$,$\angle Q = 90^{\circ}$,$PQ = 3 \, cm$,and $PR = 6 \, cm$.For $\angle PRQ$ (or $\angle R$),the side opposite is $PQ$ and the hypotenuse is $PR$.Using the trigonometric ratio: $\sin R = \frac{	ext{Opposite}}{	ext{Hypotenuse}} = \frac{PQ}{PR}$.$\sin R = \frac{3}{6} = \frac{1}{2}$.Since $\sin 30^{\circ} = \frac{1}{2}$,we have $\angle PRQ = 30^{\circ}$.In $	riangle PQR$,the sum of angles is $180^{\circ}$.$\angle QPR + \angle PQR + \angle PRQ = 180^{\circ}$.$\angle QPR + 90^{\circ} + 30^{\circ} = 180^{\circ}$.$\angle QPR = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

In $\triangle PQR$,right-angled at $Q$,$PQ = 3 \, cm$ and $PR = 6 \, cm$. Determine $\angle QPR$ and $\angle PRQ$.

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