In the figure, JKL is parallel to PQR and the line PL cuts ∠KPR into half. Given that LP and KQ are straight lines, ∠JKP = 50°, ∠PQN = 35° and ∠KNL = 116°, find
- ∠b
- ∠d
- ∠c
(a)
∠PNQ = ∠KNL = 116° (Vertically opposite angles)
∠b
= 180° - 116° - 35°
= 29° (Angles sum of triangle)
(b)
∠d
= 180° - 29°
= 151° (Interior angles)
(c)
∠c
= 180° - 29° - 29° - 35°
= 95° (Angles sum of triangle)
Answer(s): (a) 29°; (b) 151°; (c) 95°