In ΔPQR ∠Q = 66° and ∠R = 34°. T is a point on QR. and S is a point between Q and T such that PS ⊥ QR and PT is the bisector of ∠QPR. What is the measure of ∠SPT?
Answer & explanation
Correct answer: option 2
\(\angle\)QPR = \({180}^\circ\) - \({66}^\circ\) - \({34}^\circ\)
⇒ \({80}^\circ\)
\(\angle\)QPT = \(\angle\)TPR = \({40}^\circ\) [As PT angle bisector]
In \(\Delta \)PTR
\(\angle\)PTR + \(\angle\)PRT + \(\angle\)TPR = \({180}^\circ\)
⇒ \(\angle\)PTR = \({180}^\circ\) - \({34}^\circ\) - \({40}^\circ\)
⇒ \(\angle\)PTR = \({106}^\circ\)
Now,
\(\angle\)PTS + \(\angle\)PRT = \({180}^\circ\)
⇒ \(\angle\)PTS + \({106}^\circ\) = \({180}^\circ\)
⇒ \(\angle\)PTS = \({180}^\circ\) - \({106}^\circ\)
⇒ \(\angle\)PTS = \({74}^\circ\)
So,
\(\angle\)STP = \({180}^\circ\) - \({90}^\circ\) - \({74}^\circ\)
⇒ \(\angle\)SPT = \({16}^\circ\)
Therefore, \(\angle\)SPT is \({16}^\circ\).