PSLEIn the figure, GHJK is a straight line, HKLM is a parallelogram and LJ = LM. ∠MGH is a right angle, ∠GMH = 38° and ∠KLJ = 20°.
- Find ∠n.
- Find ∠p.
(a)
∠MGH = 90° (Right angle)
∠n
= 90° + 38°
= 128° (Exterior angle of a triangle)
(b)
∠KLM = ∠KHM = 128° (Parallelogram)
∠JLM
= 128° - 20°
= 108°
∠p
= (180° - 108°) ÷ 2
= 36° (Isosceles triangle)
Answer(s): (a) 128°; (b) 36°