Question

In the given figure, $\angle X=62º,\angle XYZ=54º$. If YO and ZO are the bisectors of $\angle XYZ$ and $\angle XZY$ respectively of ΔXYZ, find $\angle OZY$ and $\angle YOZ$.

Answer 1

As the sum of all interior angles of a triangle is 180º, therefore, for ΔXYZ,

$\angle X+\angle XYZ+\angle XZY=180º$

$\Rightarrow 62º+54º+\angle XZY=180º$

$\Rightarrow \angle XZY=180º-116º$

$\Rightarrow \angle XZY=64º$

$\angle OZY=$ $\frac{64}{2}$ = 32º (OZ is the angle bisector of ∠XZY)

Similarly, $\angle OYZ=\frac{54}{2}=27º$

Using angle sum property for ΔOYZ, we obtain

$\angle OYZ+\angle YOZ+\angle OZY=180º$

$\Rightarrow 27º+\angle YOZ+32º=180º$

$\Rightarrow \angle YOZ=180º-59º$

$\Rightarrow \angle YOZ=121º$

Hence, $\angle OZY={32}^0,\angle OYZ={27}^0$