Question

In the given figure line AB and CD intersect at O . $\angle BOC={36}^{\circ }$. Find $\angle x$ , $\angle y$ and $\angle z$

Answer 1

In first figure,

$\angle AOD=\angle BOC$ ------- Vertically Opposite angles

$\therefore x={36}^{\circ }$

$\angle AOC+\angle COB={180}^{\circ }$ ------ Angles on straight line

$y+{36}^{\circ }={180}^{\circ }$

$y={180}^{\circ }-{36}^{\circ }$

$y={144}^{\circ }$

$\angle AOC=\angle BOD$ ------- Vertically Opposite angles

$y=z={144}^{\circ }$

In second figure,

$\angle POR+\angle ROT+\angle QOT={180}^{\circ }$ ------ Angles on straight line

$2x+{90}^{\circ }+x={180}^{\circ }$

$3x={90}^{\circ }$

$x={30}^{\circ }$

$\therefore 2x={60}^{\circ }$

$\angle POR=\angle SOQ$ ------- Vertically Opposite angles

$2x=y={60}^{\circ }$

$\angle POS=\angle ROQ$ ------- Vertically Opposite angles

$z={90}^{\circ }+x$

$z={90}^{\circ }+{30}^{\circ }$

$z={120}^{\circ }$