You know the measure of the exterior angle which forms a linear pair with the vertex angle. Describe two ways you can find measures of the interior angles of the triangle.

Let the exterior angle be represented by [tex]x^{o}[/tex], and the three interior angles of the triangle be represented by [tex]a^{o}[/tex], [tex]b^{o}[/tex] and [tex]c^{o}[/tex].

Assume that the vertex angle [tex]c^{o}[/tex] is the linear pair to the exterior angle [tex]x^{o}[/tex].

i.e [tex]x^{o}[/tex] + [tex]c^{o}[/tex] = [tex]180^{o}[/tex] (sum of angles on a straight line)

Thus,

i. [tex]a^{o}[/tex] + [tex]b^{o}[/tex] = [tex]x^{o}[/tex] (sum of two opposite interior angles is equal to an exterior angle)

⇒ [tex]a^{o}[/tex] = [tex]x^{o}[/tex] - [tex]b^{o}[/tex]

Also,

[tex]b^{o}[/tex] = [tex]x^{o}[/tex] - [tex]a^{o}[/tex]

But,

[tex]a^{o}[/tex] + [tex]b^{o}[/tex] + [tex]c^{o}[/tex] = [tex]180^{o}[/tex] (sum of angles in a triangle)

⇒ [tex]c^{o}[/tex] = [tex]180^{o}[/tex] - ([tex]a^{o}[/tex] + [tex]b^{o}[/tex])

and

[tex]a^{o}[/tex] + [tex]b^{o}[/tex] = [tex]180^{o}[/tex] - [tex]c^{o}[/tex]

ii. [tex]a^{o}[/tex] + [tex]b^{o}[/tex] + [tex]c^{o}[/tex] = [tex]180^{o}[/tex] (sum of angles in a triangle)

[tex]a^{o}[/tex] = [tex]180^{o}[/tex] - ([tex]b^{o}[/tex] + [tex]c^{o}[/tex])

Also,

[tex]a^{o}[/tex] + [tex]b^{o}[/tex] = [tex]x^{o}[/tex]

⇒ [tex]b^{o}[/tex] = [tex]x^{o}[/tex] - [tex]a^{o}[/tex]

[tex]c^{o}[/tex] = [tex]180^{o}[/tex] - [tex]x^{o}[/tex]