In the figure, NQS and PQR are straight lines. QN = QP and RQ = RS. If ∠PNQ = 50°, find
- ∠NQR
- ∠QRS
(a)
∠NQR
= 50° + 50°
= 100° (Exterior angle of a triangle)
(b)
∠RQS
= 180° - 100°
= 80° (Angles on a straight line)
∠QRS
= 180° - 80° - 80°
= 20° (Isosceles triangle)
Answer(s): (a) 100°; (b) 20°