PSLEIn the figure, HJKL is a straight line, JLMN is a parallelogram and MK = MN. ∠NHJ is a right angle, ∠HNJ = 35° and ∠LMK = 17°.
- Find ∠p.
- Find ∠q.
(a)
∠NHJ = 90° (Right angle)
∠p
= 90° + 35°
= 125° (Exterior angle of a triangle)
(b)
∠LMN = ∠LJN = 125° (Parallelogram)
∠KMN
= 125° - 17°
= 108°
∠q
= (180° - 108°) ÷ 2
= 36° (Isosceles triangle)
Answer(s): (a) 125°; (b) 36°