Delta ABC sim Delta PQR for the correspondenc | vedclass

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Question

(A) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow QRP$.This implies that the corresponding angles are equal:$m

Vedclass · Accepted Answer

$m \angle P = 60^{\circ}, m \angle Q = 80^{\circ}, m \angle R = 40^{\circ}$

Vedclass · Answer

(A) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow QRP$.This implies that the corresponding angles are equal:$m \angle A = m \angle Q$$m \angle B = m \angle R$$m \angle C = m \angle P$Given $m \angle A = 80^{\circ}, m \angle B = 40^{\circ}, m \angle C = 60^{\circ}$.Substituting these values:$m \angle Q = 80^{\circ}$$m \angle R = 40^{\circ}$$m \angle P = 60^{\circ}$Thus,the angles of $\Delta PQR$ are $m \angle P = 60^{\circ}, m \angle Q = 80^{\circ}, m \angle R = 40^{\circ}$.

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow QRP$. If $m \angle A = 80^{\circ}, m \angle B = 40^{\circ}$ and $m \angle C = 60^{\circ}$,find the measures of all the angles of $\Delta PQR$.

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