Question

find $mangle6$.

Explanation:

Step1: Find \(m\angle3\)

Use the property that the exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. For the triangle with angles \(25^{\circ}\) and \(m\angle3\) and exterior angle \(51^{\circ}\), we have \(m\angle3=51^{\circ}-25^{\circ}\).

$$m\angle3 = 26^{\circ}$$

Step2: Find \(m\angle2\)

Since the lines are intersected in a way that we assume some right - angle or other angle relationships (not fully shown in the problem, but common in such setups). Let's assume we use the fact that in a right - angled or other standard triangle setups (if we consider the sum of angles in a triangle is \(180^{\circ}\) and some perpendiculars). But more directly, if we consider the parallel - like or congruent - triangle - like properties (visual inspection of the figure's structure). However, if we consider the last part for \(\angle6\):
We know that in a right - angled triangle (the right - angled triangle with \(\angle5\) and \(\angle6\) and the other angle). First, find \(m\angle5\).
We know that in a triangle (the large - composed triangle), if we consider the sum of angles. But more simply, for the triangle with \(\angle6\), \(\angle5\) and the right - angle (from the figure's right - angle mark). We know that \(m\angle5=28^{\circ}+35^{\circ}\) (using the exterior angle property of a non - right triangle part).

$$m\angle5=63^{\circ}$$

Step3: Find \(m\angle6\)

Since in a right - angled triangle (the triangle with \(\angle5\), \(\angle6\) and the right - angle (\(90^{\circ}\))), we use the fact that the sum of angles in a triangle is \(180^{\circ}\). So \(m\angle6 + m\angle5+90^{\circ}=180^{\circ}\), then \(m\angle6=180^{\circ}-90^{\circ}-m\angle5\). Substitute \(m\angle5 = 63^{\circ}\)

$$m\angle6=180^{\circ}-90^{\circ}-63^{\circ}=27^{\circ}$$

Answer:

\(27^{\circ}\)