Question

In a $△$ PQR,PQ=PR and $\angle Q$ is twice that of $\angle P$.Then $\angle Q$ =

• A. ${72}^0$
• B. ${36}^0$
• C. ${144}^0$
• D. ${108}^0$

Answer 1

Answer: (A)

${72}^0$

Given That:

$PQ=PR$ and $\angle Q=2\angle P$

Since, $PQ=PR\Rightarrow \angle Q=\angle R$

Now, In $△PQR$

$\angle P+\angle Q+\angle R={180}^{\circ }$ (Sum of angles of a triangle is ${180}^{\circ }$.)

$\Rightarrow \angle P+2\angle P+2\angle P={180}^{\circ }$

$\Rightarrow 5\angle P={180}^{\circ }$

$\Rightarrow P={36}^{\circ }$

$\therefore \angle Q=2\angle P=2\times {36}^{\circ }={72}^{\circ }$

Hence, A is the correct option.