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In Fig. 6.30, $\angle A=80^\circ$ and $\angle B=60^\circ$ in $\triangle ABC$. If $\triangle ABC \sim \triangle RQP$, then $\angle P$ is  
A$20^\circ$
B$40^\circ$
C$60^\circ$
D$80^\circ$
Answer & Solution
Correct answer: B. $40^\circ$
Since $\triangle ABC \sim \triangle RQP$, corresponding angles are equal, so $\angle C=\angle P$. Now $\angle C=180^\circ-\angle A-\angle B=180^\circ-80^\circ-60^\circ=40^\circ$. Hence $\angle P=40^\circ$.
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