PSLEIn the figure, HJKL is a straight line, JLMN is a parallelogram and MK = MN. ∠NHJ is a right angle, ∠HNJ = 30° and ∠LMK = 18°.
- Find ∠p.
- Find ∠q.
(a)
∠NHJ = 90° (Right angle)
∠p
= 90° + 30°
= 120° (Exterior angle of a triangle)
(b)
∠LMN = ∠LJN = 120° (Parallelogram)
∠KMN
= 120° - 18°
= 102°
∠q
= (180° - 102°) ÷ 2
= 39° (Isosceles triangle)
Answer(s): (a) 120°; (b) 39°