Delta ABC sim Delta PQR for the correspondenc | vedclass

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Question

(C) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow RPQ$. This implies that $\angle A = \angle R$,$\angle B = \a

Vedclass · Accepted Answer

$\angle P$

Vedclass · Answer

(C) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow RPQ$. This implies that $\angle A = \angle R$,$\angle B = \angle P$,and $\angle C = \angle Q$.In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.Given $m \angle A + m \angle C = m \angle B$,we substitute this into the sum equation: $m \angle B + m \angle B = 180^{\circ}$,which gives $2(m \angle B) = 180^{\circ}$,so $m \angle B = 90^{\circ}$.Since $\angle B = \angle P$ from the correspondence,it follows that $m \angle P = 90^{\circ}$.Therefore,$\angle P$ is a right angle.

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow RPQ$. If $m \angle A + m \angle C = m \angle B$,then in $\Delta PQR$,$\ldots \ldots \ldots$ is a right angle.

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