PSLEIn the figure, HJKL is a straight line, JLMN is a parallelogram and MK = MN. ∠NHJ is a right angle, ∠HNJ = 27° and ∠LMK = 13°.
- Find ∠p.
- Find ∠q.
(a)
∠NHJ = 90° (Right angle)
∠p
= 90° + 27°
= 117° (Exterior angle of a triangle)
(b)
∠LMN = ∠LJN = 117° (Parallelogram)
∠KMN
= 117° - 13°
= 104°
∠q
= (180° - 104°) ÷ 2
= 38° (Isosceles triangle)
Answer(s): (a) 117°; (b) 38°