In Fig. 6.31, if $ \mathrm{PQ} \| \mathrm{ST}, \angle \mathrm{PQR}=110^{\circ} $ and $ \angle \mathrm{RST}=130^{\circ} $, find $ \angle \mathrm{QRS} $.
[Hint : Draw a line parallel to ST through point R.]
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Given:
$PQ \parallel ST$, $\angle PQR=130^o$.

To do:

We have to find $\angle QRS$.

Solution:

Let us draw a line parallel to $ST$ through the point $R$ and name it $UV$.

We know that,

The angles on the same side of the transversal are equal to $180^o$.

Therefore,

$\angle RST+\angle SRV=180^o$

This implies,

$\angle SRV=180^o-130^o$  (Since, $\angle S=130^o$)

We get,

$\angle SRV=50^o$

In the similar way, we get,

$\angle PQR+\angle QRU=180^o$

This implies,

$\angle QRU=180^o-110^o$   (Since,$\angle Q=110^o$)

We get,

$\angle QRU=70^o$

Therefore,

$\angle QRU+\angle QRS+\angle SRV=180^o$  (Since the sum of the measures of the angles in linear pairs is always $180^o$)

Therefore by substituting the values we get,

$\angle QRS=180^o-70^o-50^o$

$\angle QRS=180^o-120^o$

$\angle QRS=60^o$

Therefore, $\angle QRS=60^o$.

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Updated on: 10-Oct-2022

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