Question

In the adjacent figure, if $QT\perp PR,\angle TQR={40}^o$ and $\angle SPR={30}^o$, find x and y.

Answer 1

In $△TQR$,
${90}^o+{40}^o+x={180}^o$ (Angle sum property of a triangle)
Therefore, $x^o={50}^o$
Now, $y^o=\angle SPR+x^o$ (Exterior angle of triangle)
Therefore, $y^o={30}^o+{50}^o$
$={80}^o$