In the figure, DEF is parallel to JKL and the line JF cuts ∠EJL into half. Given that FJ and EK are straight lines, ∠DEJ = 49°, ∠JKH = 41° and ∠EHF = 110°, find
- ∠n
- ∠q
- ∠p
(a)
∠JHK = ∠EHF = 110° (Vertically opposite angles)
∠n
= 180° - 110° - 41°
= 29° (Angles sum of triangle)
(b)
∠q
= 180° - 29°
= 151° (Interior angles)
(c)
∠p
= 180° - 29° - 29° - 41°
= 90° (Angles sum of triangle)
Answer(s): (a) 29°; (b) 151°; (c) 90°