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

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Question

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 \(\angle YOZ\)

Reasoning:

As we know the angle sum property of a triangle:

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

Steps:

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 \(\angle YOZ = 121^\circ \)

  
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