PSLEIn the figure, HJKL is a straight line, JLMN is a parallelogram and MK = MN. ∠NHJ is a right angle, ∠HNJ = 34° and ∠LMK = 22°.
- Find ∠p.
- Find ∠q.
(a)
∠NHJ = 90° (Right angle)
∠p
= 90° + 34°
= 124° (Exterior angle of a triangle)
(b)
∠LMN = ∠LJN = 124° (Parallelogram)
∠KMN
= 124° - 22°
= 102°
∠q
= (180° - 102°) ÷ 2
= 39° (Isosceles triangle)
Answer(s): (a) 124°; (b) 39°