# In Fig. 6.42, if lines $\mathrm{PQ}$ and $\mathrm{RS}$ intersect at point $\mathrm{T}$, such that $\angle \mathrm{PRT}=40^{\circ}, \angle \mathrm{RPT}=95^{\circ}$ and $\angle \mathrm{TSQ}=75^{\circ}$, find $\angle \mathrm{SQT}$."

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Given:

Lines $PQ$ and $RS$ intersect at point $T$, such that $\angle PRT=40^o, \angle RPT=95^o$ and $\angle TSQ=75^o$.

To do:

We have find $\angle SQT$.

Solution:

Let us consider $\triangle PRT$.

We know that,

The sum of the interior angles of a triangle is always $180^o$.

Therefore,

$\angle PRT+\angle RPT+\angle PTR=180^o$

By substituting the values of $\angle PRT$ and $\angle RPT$ in the above equation we get,

$95^o+40^o+\angle PTR=180^o$

$135^o+\angle PTR=180^o$

This implies,

$\angle PTR=180^o-135^o$

$\angle PTR=45^o$

We know that,

In a triangle vertically opposite angles are always equal.

This implies,

$\angle PTR=\angle STQ$

$\angle STQ=45^o$

Since the sum of the interior angles of a triangle is always $180^o$ we get,

$\angle TSQ+\angle PTR+\angle SQT=180^o$

By substituting the values we get,

$75^o+45^o+\angle SQT=180^o$

$120^o+\angle SQT=180^o$

This implies,

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

$\angle SQT=60^o$

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

Updated on 10-Oct-2022 13:40:46