Question

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Explanation:

Step1: Find the value of \(a\)

Since the sum of an angle and its adjacent supplementary angle is \(180^{\circ}\), for the angle \(116^{\circ}\), we have \(a = 180^{\circ}- 116^{\circ}\)
\(a=64^{\circ}\)

Step2: Use the property of exterior - angle sum of a polygon (Here, we can consider the angles related to the non - parallel lines and the constructed right - angle)

Let's assume the polygon - like figure formed by the angles. The sum of the exterior angles of a polygon is \(360^{\circ}\). But here, we can also use the property of angle - chasing.
We know one angle is \(82^{\circ}\), another is \(90^{\circ}\) (right - angle), and we found \(a = 64^{\circ}\). Let the angle related to \(b\) be considered.
First, we find the angle adjacent to \(82^{\circ}\) which is \(180^{\circ}-82^{\circ}=98^{\circ}\)
The sum of angles around the "vertex" (using the fact that the sum of angles in a "loose" polygon - like structure with non - overlapping angles around a set of lines) is \(360^{\circ}\).
Let's use another approach:
We know that the sum of angles in a quadrilateral - like figure (formed by the non - parallel lines and the right - angle) is \(360^{\circ}\). But we can also use the property of alternate - segment - like angle relations (by constructing parallel lines conceptually).
Let's first find the angle adjacent to the \(90^{\circ}\) - related part.
We know that \(b=\frac{1}{2}(360^{\circ}-90^{\circ}-(180 - 82)^{\circ}-(180 - 116)^{\circ})\)
First, calculate \(180 - 82=98^{\circ}\) and \(180 - 116 = 64^{\circ}\)
\(360^{\circ}-90^{\circ}-98^{\circ}-64^{\circ}=108^{\circ}\)
Since the two angles (the ones with the same marking) are equal (isosceles - like property as the segments are marked equal), \(b=\frac{1}{2}\times108^{\circ}\)

Answer:

\(a = 64^{\circ}\), \(b = 54^{\circ}\)