Question

In the given figure, if $y={32}^o$ and $z={40}^o$, then determine $x$. If $y+z={90}^o$, then prove that $x={45}^o$.

Answer 1

Let $\angle BAO=a$

Hence $OCB={180}^{\circ }-a$

Now, $\angle OCD=a$

In $△OCD$

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

Now,$\angle OAE={180}^{\circ }a$

In$△AOE$

$180-a+x+x={180}^{\circ }$

$⟹x+z-a={180}^{\circ }-{180}^{\circ }$

$⟹x+z=a$

Now, $x+x+x+y={180}^{\circ }$

$⟹2x+x+y={180}^{\circ }$

As it is given $y+x={90}^{\circ }$

$⟹2x={90}^{\circ }$

$x={45}^{\circ }$

Henc proved.