Question

In the given figure, $AOB$ is a diameter of the circle with centre $O$. $CD$ is the $\perp$ bisector of $OA$. Which of the angles $x,y,z$ is/are equal to ${30}^o$

• A. $x$ only
• B. $x$ and $y$ only
• C. $x,y,z$
• D. all of them

Answer 1

The correct option is C $x,y,z$

In the semicircle $ABC$
$x+y+z=90°⟶\left(1\right)$ ($\angle$ is semicircle)
$y+z+w=90°⟶\left(2\right)$ (sum of $\angle$ in $△)$
from (1), we get $2y+z=90°⟶\left(3\right)(\because x=y)$
from (2), we get $y+z=90°⟶\left(4\right)$
Solving (3) $\&$ (4), we get $x=30°,w=30°$ and all are 30°.