In the figure, NQS and PQR are straight lines. QN = QP and RQ = RS. If ∠PNQ = 52°, find
- ∠NQR
- ∠QRS
(a)
∠NQR
= 52° + 52°
= 104° (Exterior angle of a triangle)
(b)
∠RQS
= 180° - 104°
= 76° (Angles on a straight line)
∠QRS
= 180° - 76° - 76°
= 28° (Isosceles triangle)
Answer(s): (a) 104°; (b) 28°