Question

the angle bisectors of \\( \triangle j k l \\) are \\( \overline{j q}, \overline{k q} \\), and \\( \overline{l q} \\). they meet at a single point \\( q \\).
(in other words, \\( q \\) is the incenter of \\( \triangle j k l \\).)
suppose \\( p q=17, k q=18, m \angle m j p=92^{circ} \\), and \\( m \angle n k q=24^{circ} \\).
find the following measures.
note that the figure is not drawn to scale.

Explanation:

Step1: Find \(m\angle NKM\)

Since \(KQ\) is an angle - bisector, \(m\angle NKM = 2m\angle NKQ\).
Given \(m\angle NKQ = 24^{\circ}\), then \(m\angle NKM=2\times24^{\circ}=48^{\circ}\).

Step2: Find \(MQ\)

Since \(Q\) is the in - center of \(\triangle JKL\), the in - center is equidistant from the sides of the triangle. So \(MQ = PQ\).
Given \(PQ = 17\), then \(MQ = 17\).

Step3: Find \(m\angle NLQ\)

First, since \(JQ\) is an angle - bisector, \(m\angle MJP = 2m\angle M JQ\). Given \(m\angle MJP=92^{\circ}\), then \(m\angle MJQ=\frac{92^{\circ}}{2}=46^{\circ}\).
In \(\triangle JKL\), the sum of angles in a triangle is \(180^{\circ}\). Let's assume we know two angles from the angle - bisector information.
We know \(m\angle K = 48^{\circ}\) (from step 1) and \(m\angle J = 92^{\circ}\) (after using the angle - bisector property for \(\angle J\)). Then \(m\angle L=180-(92 + 48)=40^{\circ}\).
Since \(LQ\) is an angle - bisector, \(m\angle NLQ=\frac{1}{2}m\angle L\). So \(m\angle NLQ=\frac{40^{\circ}}{2}=20^{\circ}\).

Answer:

\(m\angle NKM = 48^{\circ}\), \(MQ = 17\), \(m\angle NLQ = 20^{\circ}\)