In Fig. 6.14, lines $ \mathrm{XY} $ and $ \mathrm{MN} $ intersect at $ O $. If $ \angle \mathrm{POY}=90^{\circ} $ and $ a: b=2: 3 $, find $ c $.
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Given:
 Lines $XY, MN$ intersect at $O$ , $\angle POY=90^o$ and $a:b=2:3$.

 To do: 

We have to find $c$.

Solution:

Given,

$\angle POY=90^o$ and $a:b=2:3$

We know that,

The sum of the measures of the angles in linear pairs is always $180^o$.

This implies,

$\angle POY+a+b=180^o$

By substituting $\angle POY=90^o$ in the above equation 

We get,

$90^o+a+b=180^o$

$a+b=180^o-90^o$

$a+b=90^o$

Let $a$ be $2x$ and $b$ be $3x$ (since, $a:b=2:3$)

This implies,

$2x+3x=90^o$

$5x=90^o$

$x=\frac{90^o}{5}$

$x=18^o$

Therefore,

$a=2\times18^o$

$a=36^o$ and

$b=3\times18^o$

$b=54^o$

Similarly, as $b$ and $c$ are also in straight line

We get,

$b+c=180^o$

This implies,

$54^o+c=180^o$

$c=180^o-54^o$

$c=126^o$

Therefore, $c=126^o$.