Question

In the figure, QT $\perp$ PR, $\angle TQR={40}^o$ and $\angle SPR={30}^o$. Find $\angle TRS$ and $\angle PSQ$.

Answer 1

Consider right $△QTR$ ,

Where, $\angle T={90}^o$ [$QT\perp PR$]

Sum of all the angles of a triangle is ${180}^o$

Hence, $\angle RQT+\angle QTS+\angle TRQ={180}^o$

$\Rightarrow {40}^o+{90}^o+\angle R={180}^o$

$\angle TRS={180}^o-{130}^o={50}^o$

Now, $\angle PSQ=\angle SPR+\angle PRS$ (Exterior angle is equal to sum of interior opposite angles.)

$\angle PSQ={30}^o+{50}^o={80}^o$