Question

first, find ( mangle jqk ).
( mangle jqk=square^{circ} )

Explanation:

Step1: Use the property of vertical angles

Vertical angles are equal. The sum of angles around a point \(Q\) is \(180^{\circ}\). We know that \(\angle JQP = 80^{\circ}\) and \(\angle KQL=30^{\circ}\). Also, \(\angle JQK+\angle KQL+\angle LQM+\angle MQN+\angle NQP+\angle PJQ = 360^{\circ}\), but since \(\angle PJQ\) and \(\angle LQM\) are vertical angles (\(\angle PJQ=\angle LQM = 80^{\circ}\)), \(\angle KQL\) and \(\angle NQM\) are vertical angles (\(\angle KQL=\angle NQM = 30^{\circ}\)), and \(\angle JQK\) and \(\angle MQP\) are vertical angles. Another way is to use the fact that \(\angle JQK + 80^{\circ}+30^{\circ}=180^{\circ}\) (because they form a linear - pair with the opposite angles).

Step2: Solve for \(\angle JQK\)

We have the equation \(m\angle JQK+80^{\circ}+30^{\circ}=180^{\circ}\). Subtract \(80^{\circ}+30^{\circ}\) from both sides. \(m\angle JQK=180^{\circ}-(80^{\circ} + 30^{\circ})\).

$$m\angle JQK=180^{\circ}-110^{\circ}=70^{\circ}$$

Answer:

\(70\)