Question

If $m\angle \text{OAB}={50}^{\circ }$, then sum of angles $x,y$ and $z$ is

• A. ${170}^{\circ }$
• B. ${160}^{\circ }$
• C. ${150}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is A ${170}^{\circ }$

Let's find the value of $x,y$ and $z$
$\Rightarrow x={50}^{\circ }$
($\because △\text{AOB}$ is an isoceles)
Hence,
$x+y+{50}^{\circ }={180}^{\circ }$
$y={180}^{\circ }-x-{50}^{\circ }$
$\Rightarrow y={80}^{\circ }$
Now,
$z=\frac{y}{2}$
(The $\angle \text{AOB}$ subtended by the arc $\text{AB}$ at the center is double the $\angle \text{BCA}=z$, subtended by it at point $\text{C}$ on the remaining part of the circle.)
$\Rightarrow z=\frac{{80}^{\circ }}{2}$
$={40}^{\circ }$
So,
$\Rightarrow x+y+z={50}^{\circ }+{80}^{\circ }+{40}^{\circ }$
$={170}^{\circ }$