Question

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

Answer 1

Since $\overline{ AB } \| \overline{ PQ }, AP = BQ$
$\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} $
$\therefore \angle AOQ +\angle ACQ =90^{\circ}+45^{\circ}=135^{\circ}$