Question

In figure, rays AB and CD intersect at O.


ii. Determine $x$ when $y={40}^{\circ }$

Answer 1

$\text{Given}:y={40}^{\circ },\angle AOC=2x\text{and}\angle BOC=y$

As, AB is a straight line
$\therefore \angle AOC+\angle BOC={180}^{\circ }\left(\text{Linear pair}\right)$

$2x+y={180}^{\circ }$

$2x+{40}^{\circ }={180}^{\circ }$

$2x={180}^{\circ }-{40}^{\circ }$

$2x={140}^{\circ }$

$x=\frac{{140}^{\circ }}{2}$

$x={70}^{\circ }$