Question

What value of y would make AOB, a line in the figure, if $\angle AOC=4y$ and $\angle BOC=(6y+30)$.

Answer 1

In the figure, AOB is a line if
$\angle AOC+\angle BOC={180}^{\circ }$
$\Rightarrow 6y+{30}^{\circ }+4y={180}^{\circ }\Rightarrow 10y={180}^{\circ }-{30}^{\circ }={150}^{\circ }\Rightarrow y=\frac{{150}^{\circ }}{10}={15}^{\circ }\therefore y={15}^{\circ }$