Question

In the below figure, $\mathrm{\angle }X={62}^{\circ }$, $\mathrm{\angle }XYZ={54}^{\circ }$. If YO and ZO are the bisectors of angle Y and angle Z then find $\mathrm{\angle }OZY\mathrm{\&}\mathrm{\angle }YOZ$.

Answer 1

Hint: We know that the sum of all the angles of a triangle are ${180}^{\circ }$ so using this property we can find the value of the third angle Z of the triangle XYZ because we know the values of angle X and angle Y. After that it is given that OY and OZ are the angle bisectors of angles Y and Z respectively so $\mathrm{\angle }OYZ\mathrm{\&}\mathrm{\angle }OZY$ are half of angles Y and Z using this information we can find the value of $\mathrm{\angle }OZY$ and $\mathrm{\angle }YOZ$ is calculated by using the property of the triangle that sum of the all the angles of a triangle are ${180}^{\circ }$ and here the triangle is OYZ.

Complete step by step answer:
We have given a figure below, in which $\mathrm{\angle }X={62}^{\circ }$, $\mathrm{\angle }XYZ={54}^{\circ }$.

We are asked to find $\mathrm{\angle }OZY\mathrm{\&}\mathrm{\angle }YOZ$.
First of all we are going to find the value of angle Z which we do by using the property of the triangle of XYZ that the sum of all angles of the triangle are ${180}^{\circ }$.
$\mathrm{\angle }X+\mathrm{\angle }Y+\mathrm{\angle }Z={180}^{\circ }$
Substituting $\mathrm{\angle }X={62}^{\circ }$ and $\mathrm{\angle }Y={54}^{\circ }$ in the above equation we get,
$\begin{array}{rl}& {62}^{\circ }+{54}^{\circ }+\mathrm{\angle }Z={180}^{\circ }.......Eq.(1)\\ & \Rightarrow {116}^{\circ }+\mathrm{\angle }Z={180}^{\circ }\\ & \Rightarrow \mathrm{\angle }Z={180}^{\circ }-{116}^{\circ }\\ & \Rightarrow \mathrm{\angle }Z={64}^{\circ }\end{array}$
Hence, we got the value of angle Z as ${64}^{\circ }$.
It is also given that OY and OZ are the angle bisectors of angles Y and Z respectively. This means that the values of $\mathrm{\angle }OYZ\mathrm{\&}\mathrm{\angle }OZY$ are equal to half of angle Y and half of angle Z respectively.
$\begin{array}{rl}& \mathrm{\angle }OYZ={\displaystyle \frac{1}{2}}\mathrm{\angle }Y\\ & \Rightarrow \mathrm{\angle }OYZ={\displaystyle \frac{1}{2}}\left({54}^{\circ }\right)\\ & \Rightarrow \mathrm{\angle }OYZ={27}^{\circ }\end{array}$
From the above, we got the value of $\mathrm{\angle }OYZ={27}^{\circ }$.
$\begin{array}{rl}& \mathrm{\angle }OZY={\displaystyle \frac{1}{2}}\mathrm{\angle }Z\\ & \Rightarrow \mathrm{\angle }OZY={\displaystyle \frac{1}{2}}\left({64}^{\circ }\right)\\ & \Rightarrow \mathrm{\angle }OZY={32}^{\circ }\end{array}$
From the above, we got the value of $\mathrm{\angle }OZY={32}^{\circ }$.
Now, to find the value of $\mathrm{\angle }YOZ$ we are going to use the property of triangle which says that the sum of the angles of a triangle are ${180}^{\circ }$ in $\mathrm{\Delta }YOZ$.
$\begin{array}{rl}& \mathrm{\angle }YOZ+\mathrm{\angle }OZY+\mathrm{\angle }OYZ={180}^{\circ }\\ & \Rightarrow \mathrm{\angle }YOZ+{32}^{\circ }+{27}^{\circ }={180}^{\circ }\\ & \Rightarrow \mathrm{\angle }YOZ+{59}^{\circ }={180}^{\circ }\\ & \Rightarrow \mathrm{\angle }YOZ={180}^{\circ }-{59}^{\circ }\\ & \Rightarrow \mathrm{\angle }YOZ={121}^{\circ }\end{array}$
From the above, we got the value of $\mathrm{\angle }YOZ={121}^{\circ }$.

Hence, we got the value of $\mathrm{\angle }YOZ={121}^{\circ }$ and $\mathrm{\angle }OZY={32}^{\circ }$.

Note: This question demands the knowledge of the property of a triangle which says that sum of all angles of the triangle are ${180}^{\circ }$ if you don’t know this property then you are lost in this problem and you cannot solve anything in it.
The mistake that could happen in this problem is the calculation mistakes so be careful. Another thing through which you can check whether the angles that you calculated are correct or not by using these angles in the equation of sum of all the angles of the triangle as ${180}^{\circ }$.
For e.g. to check whether $\mathrm{\angle }OZY={32}^{\circ }$, first of all double it so that we will get the angle Z.
$2\mathrm{\angle }OZY={64}^{\circ }=\mathrm{\angle }Z$
Now, substituting the above value of angle Z in eq. (1) we get,
$\begin{array}{rl}& {62}^{\circ }+{54}^{\circ }+{64}^{\circ }={180}^{\circ }\\ & \Rightarrow {180}^{\circ }={180}^{\circ }\end{array}$
As you can see that L.H.S is equal to R.H.S so the value of $\mathrm{\angle }OZY={32}^{\circ }$ which we have calculated is correct.