Draw a figure representing that situation:
Since the angle A and the angle of 68° are supplementary angles, then:
[tex]A+68=180[/tex]Substract 68 from both sides of the equation to find A:
[tex]A=112[/tex]Since the sum of the interior angles of a triangle is always 180, then:
[tex]\begin{gathered} A+B+B=180 \\ \Rightarrow A+2B=180 \end{gathered}[/tex]Substitute for A=112:
[tex]112+2B=180[/tex]Substract 112 from both sides of the equation:
[tex]2B=68[/tex]Divide both sides of the equation by 2:
[tex]B=34[/tex]Therefore, the measure of both remote angles is 34°.
(Both have the same measure, since it is an isosceles triangle).
