In the given figure, if lines $P Q$ and $R S$ intersect at point $T$, such that $\angle P R T=40^{\circ}, \angle R P T=95^{\circ}$ and $\angle T S Q=75^{\circ}$, find $\angle S Q T$.

Solution:

Using angle sum property for $\triangle \mathrm{PRT}$, we obtain

$\angle \mathrm{PRT}+\angle \mathrm{RPT}+\angle \mathrm{PTR}=180^{\circ}$

$40^{\circ}+95^{\circ}+\angle \mathrm{PTR}=180^{\circ}$

$\angle P T R=180^{\circ}-135^{\circ}$

$\angle \mathrm{PTR}=45^{\circ}$

$\angle S T Q=\angle P T R=45^{\circ}$ (Vertically opposite angles)

$\angle S T Q=45^{\circ}$

By using angle sum property for $\triangle S T Q$, we obtain

$\angle S T Q+\angle S Q T+\angle Q S T=180^{\circ}$

$45^{\circ}+\angle S Q T+75^{\circ}=180^{\circ}$

$\angle S Q T=180^{\circ}-120^{\circ}$

$\angle S Q T=60^{\circ}$