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 = 42°, ∠PQN = 37° and ∠KNL = 114°, find
- ∠b
- ∠d
- ∠c
(a)
∠PNQ = ∠KNL = 114° (Vertically opposite angles)
∠b
= 180° - 114° - 37°
= 29° (Angles sum of triangle)
(b)
∠d
= 180° - 29°
= 151° (Interior angles)
(c)
∠c
= 180° - 29° - 29° - 37°
= 101° (Angles sum of triangle)
Answer(s): (a) 29°; (b) 151°; (c) 101°