Answer:
Step-by-step explanation:
[tex]\large\text{I will label the figure so that you don't get confused.}[/tex]
[tex]\large\text{b)}[/tex]
[tex]\large\text{are supplementary because they are the angles of straight a}[/tex]
[tex]\large\text{line.}[/tex]
[tex]\large\text{$z+50^\circ=180^\circ$}[/tex]
[tex]\large\text{$z=180^\circ-50^\circ=130^\circ$}[/tex]
[tex]\large\text{base angles of isosceles triangle OBC.}[/tex]
[tex]\large\text{$\angle$OBC = $\angle$OCB = $y$}[/tex]
[tex]\large\text{angles and the opposite exterior angles to these angles is}[/tex]
[tex]\large\text{$\angle$AOC. The angle AOC will be equal to the sum of these}[/tex]
[tex]\large\text{interior opposite angles.}[/tex]
[tex]\large\text{$\angle$AOC $=\angle$OBC + $\angle$OCB}[/tex]
[tex]\large\text{$50^\circ=y+y$}[/tex]
[tex]\large\text{$50^\circ=2y$}[/tex]
[tex]\large\text{$y=25^\circ$}[/tex]
Here's another simple way to find the value of angle y if you're confused:
[tex]\large\text{the triangle OBC. Hence, their sum equals $180^\circ$.}[/tex]
[tex]\large\text{$\angle$BOC + $\angle$OBC + $\angle$OCB = $180^\circ$}[/tex]
[tex]\large\text{$z+y+y=180^\circ$}[/tex]
[tex]\large\text{$130^\circ+2y=180^\circ$}[/tex]
[tex]\large\text{$2y=180^\circ-130^\circ$}[/tex]
[tex]\large\text{$2y=50^\circ$}[/tex]
[tex]\large\text{$y=25^\circ$}[/tex]
