Question

These are numeric entry type questions.
In the given figure, ' $O^{\prime }$ is the centre of the circle. $△ABC$ and $△OPQ$ are equilateral triangles. $\overline{AB}$ is parallel to $\overline{PQ}$. Then $\angle OQA+$ $\angle ACQ=$______

Answer 1

Answer: 135

Since $\overline{AB}\Vert \overline{PQ},AP=BQ$
$\text{ Since }\angle ACB=60^{\circ },\angle AOB=120^{\circ }\text{, and }\angle PO$
$=60^{\circ }\text{, }$
$\angle AOP+\angle QOB=60^{\circ }\Rightarrow \angle AOP=\angle QOB$
$=30^{\circ }(\because AD=BE)$
$\angle AOQ=30^{\circ }+60^{\circ }=90^{\circ }$
$\angle ACQ=\frac{90^{\circ }}{2}=45^{\circ }$
$\therefore \angle AOQ+\angle ACQ=90^{\circ }+45^{\circ }=135^{\circ }$