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(A) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.Given $m \angle A + m \angle B = 130^{\circ}$,substituti

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$55$

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(A) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.Given $m \angle A + m \angle B = 130^{\circ}$,substituting this into the sum gives $130^{\circ} + m \angle C = 180^{\circ}$,so $m \angle C = 50^{\circ}$.Given $m \angle B + m \angle C = 125^{\circ}$,substituting $m \angle C = 50^{\circ}$ gives $m \angle B + 50^{\circ} = 125^{\circ}$,so $m \angle B = 75^{\circ}$.Since $m \angle A + m \angle B = 130^{\circ}$,we have $m \angle A + 75^{\circ} = 130^{\circ}$,so $m \angle A = 55^{\circ}$.Given the correspondence $ABC \leftrightarrow QPR$,the corresponding angles are $\angle A \leftrightarrow \angle Q$,$\angle B \leftrightarrow \angle P$,and $\angle C \leftrightarrow \angle R$.Therefore,$m \angle Q = m \angle A = 55^{\circ}$.

If $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow QPR$,and given $m \angle A + m \angle B = 130^{\circ}$ and $m \angle B + m \angle C = 125^{\circ}$,find $m \angle Q$. (in $^{\circ}$)

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