Question

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

Explanation:

Step1: Sum of angles at a point

Angles around point Q total $360^\circ$.

Step2: Calculate unknown angle

Let $m\angle JQK = x$. Given angles: $80^\circ$, $30^\circ$, and two equal angles (vertical angles: $\angle JQN \cong \angle KQM$, $\angle PQJ \cong \angle LQM$). Wait, correction: Assume $\angle JQK$ is the angle between JQ and KQ. Wait, standard: angles around Q: let’s denote the angle between JQ and PQ as $80^\circ$, between KQ and LQ as $30^\circ$. The remaining angles: let $x = m\angle JQK$. Since vertical angles are equal, but actually, the correct way: if we consider the straight lines? No, all lines meet at Q. Wait, the key is: the angle between JQ and KQ is the unknown. Wait, no—wait, the problem says "find $m\angle JQK$". From the diagram, the angle between JQ and KQ is equal to the angle opposite? No, wait, let's count: the total around Q is $360^\circ$. Suppose the angles are: $\angle PJQ = 80^\circ$, $\angle KQL = 30^\circ$, and $\angle JQK = x$, with the other three angles (vertical to these or complementary?) Wait, no—actually, the correct assumption here (common problem) is that the angle between JQ and KQ is $50^\circ$? Wait, no: wait, $180^\circ - 80^\circ - 30^\circ = 70^\circ$? No, wait, let's do it properly:

Wait, the diagram shows lines meeting at Q: let’s say line JL and line KM intersect? No, all lines are from Q. The angle between PQ and JQ is $80^\circ$, between KQ and LQ is $30^\circ$. The angle between JQ and KQ is $x$. Since the sum of angles on one side of a line is $180^\circ$? No, all angles around Q sum to $360^\circ$. Wait, no—wait, the problem is likely a standard one where $\angle JQK = 180^\circ - 80^\circ - 30^\circ = 70^\circ$? No, wait, $80 + 30 + x = 180$ (if they are on a straight line?), no. Wait, no—wait, the correct answer for this common problem is $50^\circ$? Wait, no, let's think again:

Wait, the angle between JQ and KQ is equal to $180^\circ - 80^\circ - 50^\circ$? No, I think I made a mistake. Wait, the standard problem here: the angle between JQ and KQ is $50^\circ$? No, wait, let's do it step by step correctly:

Total angles around Q: $360^\circ$. Suppose there are 6 angles: $80^\circ$, $x$, $30^\circ$, and their vertical angles (each pair equal). So $2(80 + x + 30) = 360$. Then $80 + x + 30 = 180$, so $x = 180 - 110 = 70$. Yes! That's it. Because vertical angles are equal, so the three distinct angles are $80^\circ$, $x$, $30^\circ$, each appearing twice. So $2*(80 + x + 30) = 360$.

Step1: Use vertical angles property

Angles around Q: 3 pairs of vertical angles.

Step2: Set up equation

$2(80^\circ + m\angle JQK + 30^\circ) = 360^\circ$

Step3: Solve for $m\angle JQK$

$80 + m\angle JQK + 30 = 180$ → $m\angle JQK = 180 - 110 = 70$

Answer:

70