If Delta PQR sim Delta XYZ under the correspo | vedclass

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(B) In $\Delta PQR$,the sum of angles is $180^{\circ}$.$m \angle P + m \angle Q + m \angle R = 180^{\circ}$.$80^{\circ} + 40^{\circ} + m \angle R = 18

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) In $\Delta PQR$,the sum of angles is $180^{\circ}$.$m \angle P + m \angle Q + m \angle R = 180^{\circ}$.$80^{\circ} + 40^{\circ} + m \angle R = 180^{\circ}$.$120^{\circ} + m \angle R = 180^{\circ} \implies m \angle R = 60^{\circ}$.Given the correspondence $PQR \leftrightarrow YZX$,the corresponding angles are equal:$m \angle P = m \angle Y = 80^{\circ}$,$m \angle Q = m \angle Z = 40^{\circ}$,$m \angle R = m \angle X = 60^{\circ}$.Therefore,$m \angle Z = m \angle Q = 40^{\circ}$.

If $\Delta PQR \sim \Delta XYZ$ under the correspondence $PQR \leftrightarrow YZX$,and $m \angle P = 80^{\circ}$ and $m \angle Q = 40^{\circ}$,find $m \angle Z$. (in $^{\circ}$)

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