Question

In the given figure $OA,OB$ are opposite rays and $\angle$ $AOC$ $+$ $\angle$ $BOD$ $=$ ${90}^0$, then $\angle$ $COD$ is

• A. ${90}^0$
• B. ${60}^0$
• C. ${45}^0$
• D. ${30}^0$

Answer 1

The correct option is A ${90}^0$

We know that sum of angles along a straight line is ${180}^0$.

Now, we can write:

$\angle AOC$$+$$\angle BOD$$+$$\angle COD$$=$${180}^0$

$⟹$ $\angle COD$$=$${180}^0$$-$$(\angle AOC+\angle BOD)$

$\because$ $\angle AOC$$+$$\angle BOD$$=$${90}^0$

$⟹$ $\angle COD$$=$${180}^0-{90}^0$

$⟹$ $\angle COD$$=$${90}^0$