Ex.6.3 Q2 Lines and Angles Solution - NCERT Maths Class 9


In Fig. below, \(\angle X = 62^\circ ,{\rm{ }}\angle XYZ = 54^\circ \). If \(YO\) and \(ZO\) are the bisectors of \(\angle XYZ\) and \(\angle XZY\) respectively of \(\Delta XYZ,\) find \(\angle OZY\) and \(\angle YOZ\) .

Text Solution

What is known?

\(\angle X = 62^\circ ,{\rm{ }}\angle XYZ = 54^\circ \) and \(YO\) and \(ZO\) are the bisectors of \(\angle XYZ\) and \(\angle XZY\) respectively.

What is unknown?

\(\angle OZY\) and \(\angle YOZ\)


As we know the angle sum property of a triangle:

Sum of the interior angles of a triangle is \(360^\circ\).


Given, in \(\Delta XYZ\),

\[\begin{align}\angle X &= 62^\circ \\\angle XYZ &= 54^\circ \end{align}\]

\(\begin{align}\angle X + \angle XYZ + \angle Z &= 180^\circ\\ ( \text{Angle sum property of }&\text{a triangle}\rm{.} )\\\\62^\circ + {\rm{ 5}}4^\circ + \angle Z &= 180^\circ \\\angle Z &=180^\circ-116^\circ \\\angle Z &= 64^\circ \end{align}\)

Now, \(OZ\) is angle bisector of \(\angle XZY\)

\(\begin{align} \angle OZY = \frac{1}{2}\angle XZY &= \frac{1}{2} \times 64^\circ \\&= 32^\circ {\rm{   }}\left( {\rm{i}} \right)\end{align} \)

Similarly, \(OY\) is angle bisector of \(\angle XYZ\)

\(\begin{align} \angle OYZ = \frac{1}{2}\angle XYZ &= \frac{1}{2} \times 54^\circ \\ &= 27^\circ \left( {{\rm{ii}}} \right)\end{align} \)

Now, in \(\Delta OYZ\)

\(\begin{align}\angle OYZ \!+\! \angle OZY \!+\! \angle YOZ &= 180^\circ \\( \text{Angle sum property of a }&\text{triangle}\rm{.} )\\\\27^\circ +32^\circ + \angle YOZ & = 180^\circ \\\left[ {{\textrm{from }}\left( {\rm{i}} \right){\rm{ and }}\left( {{\rm{ii}}} \right)} \right]\\\\\angle YOZ &\!= 180^\circ \!- 59^\circ \\\angle YOZ &= 121^\circ \end{align}\)

Hence, \(\angle OZY = 32^\circ \) and \(\angle YOZ = 121^\circ \)

Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school