Question

from the information in the diagram, what is m∠h?

Explanation:

Step1: Determine the congruent triangles

By the Side - Angle - Side (SAS) congruence criterion, \(\triangle HSN\cong\triangle KBA\).

Step2: Use the angle - sum property of triangles

In \(\triangle KBA\), we know that the sum of angles in a triangle is \(180^{\circ}\). Let \(m\angle B = m\angle S\), \(m\angle H=m\angle K\), \(m\angle N = 102^{\circ}\), \(m\angle A=48^{\circ}\).
We use the formula \(m\angle K + m\angle B+m\angle A=180^{\circ}\). Since \(m\angle N = 102^{\circ}\), and \(m\angle N+m\angle S + m\angle H=180^{\circ}\) (angle - sum property of \(\triangle HSN\)), and because of congruence \(m\angle S=m\angle B\), \(m\angle H=m\angle K\).
First, find \(m\angle K\) in \(\triangle KBA\): \(m\angle K=180^{\circ}-(m\angle B + m\angle A)\). But from the congruent triangles, we can also directly use the fact that in \(\triangle HSN\), \(m\angle H=180^{\circ}-m\angle N - m\angle S\). Since \(m\angle S=m\angle B\) and from the relationship of the two congruent triangles (by SAS, as two sides and the included angle are equal), we can calculate \(m\angle H\) as follows:
We know that \(m\angle H=180^{\circ}-102^{\circ}- 30^{\circ}\).
Another way: Since the two triangles are congruent (by SAS, as the markings show two pairs of equal sides and the included angles' relationship), in \(\triangle HSN\), \(m\angle H=180^{\circ}-102^{\circ}-48^{\circ}\) (because of the angle - sum property of triangles \(180^{\circ}\) and the correspondence from congruence).

$$m\angle H=180^{\circ}-102^{\circ}-48^{\circ}=30^{\circ}$$

Answer:

\(30\)