If Delta ABC sim Delta PQR under the correspo | vedclass

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(C) Given that $\Delta ABC \sim \Delta RPQ$. By the property of similar triangles,the corresponding angles are equal. The correspondence is $A \leftri

Vedclass · Accepted Answer

$95$

Vedclass · Answer

(C) Given that $\Delta ABC \sim \Delta RPQ$. By the property of similar triangles,the corresponding angles are equal. The correspondence is $A \leftrightarrow R$,$B \leftrightarrow P$,and $C \leftrightarrow Q$. Therefore,$m\angle A = m\angle R = 35^{\circ}$ and $m\angle B = m\angle P = 50^{\circ}$. In $\Delta ABC$,the sum of angles is $180^{\circ}$. $m\angle C = 180^{\circ} - (m\angle A + m\angle B) = 180^{\circ} - (35^{\circ} + 50^{\circ}) = 180^{\circ} - 85^{\circ} = 95^{\circ}$. Since $C \leftrightarrow Q$,we have $m\angle C = m\angle Q$. Thus,$m\angle Q = 95^{\circ}$.

If $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow RPQ$,and $m\angle A = 35^{\circ}$ and $m\angle B = 50^{\circ}$,find $m\angle Q$. (in $^{\circ}$)

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