In Delta ABC and Delta PQR,m angle A = 30^{ci | vedclass

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Question

(D) In $\Delta ABC$,the sum of angles is $180^{\circ}$.$m \angle B = 180^{\circ} - m \angle A - m \angle C = 180^{\circ} - 30^{\circ} - 60^{\circ} = 9

Vedclass · Accepted Answer

$QPR$

Vedclass · Answer

(D) In $\Delta ABC$,the sum of angles is $180^{\circ}$.$m \angle B = 180^{\circ} - m \angle A - m \angle C = 180^{\circ} - 30^{\circ} - 60^{\circ} = 90^{\circ}$.Now,comparing $\Delta ABC$ and $\Delta PQR$:$m \angle A = 30^{\circ}$ and $m \angle Q = 30^{\circ}$,so $m \angle A = m \angle Q$.$m \angle B = 90^{\circ}$ and $m \angle P = 90^{\circ}$,so $m \angle B = m \angle P$.By the $AA$ similarity criterion,the vertices must correspond in the order of their equal angles.Therefore,the correspondence $ABC \leftrightarrow QPR$ is a similarity.

In $\Delta ABC$ and $\Delta PQR$,$m \angle A = 30^{\circ}$,$m \angle C = 60^{\circ}$,$m \angle P = 90^{\circ}$ and $m \angle Q = 30^{\circ}$. Then,the correspondence $ABC \leftrightarrow \ldots$ is a similarity.

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