Given Delta ABC sim Delta PQR for the corresp | vedclass

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(D) Since $\Delta ABC \sim \Delta PQR$,the corresponding angles are equal: $m \angle A = m \angle P$,$m \angle B = m \angle Q$,and $m \angle C = m \an

Vedclass · Accepted Answer

$32$

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(D) Since $\Delta ABC \sim \Delta PQR$,the corresponding angles are equal: $m \angle A = m \angle P$,$m \angle B = m \angle Q$,and $m \angle C = m \angle R = 100^{\circ}$.In $\Delta ABC$,the sum of angles is $180^{\circ}$,so $m \angle A + m \angle B + m \angle C = 180^{\circ}$.Substituting $m \angle C = 100^{\circ}$,we get $m \angle A + m \angle B = 80^{\circ}$.Given $2 m \angle P = 3 m \angle Q$,and since $m \angle P = m \angle A$ and $m \angle Q = m \angle B$,we have $2 m \angle A = 3 m \angle B$,which implies $m \angle A = 1.5 m \angle B$.Substitute this into $m \angle A + m \angle B = 80^{\circ}$:$1.5 m \angle B + m \angle B = 80^{\circ}$$2.5 m \angle B = 80^{\circ}$$m \angle B = 80 / 2.5 = 32^{\circ}$.

Given $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. If $2 m \angle P = 3 m \angle Q$ and $m \angle C = 100^{\circ}$,find $m \angle B$. (in $^{\circ}$)

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