PSLEIn the figure, GHJK is a straight line, HKLM is a parallelogram and LJ = LM. ∠MGH is a right angle, ∠GMH = 34° and ∠KLJ = 14°.
- Find ∠n.
- Find ∠p.
(a)
∠MGH = 90° (Right angle)
∠n
= 90° + 34°
= 124° (Exterior angle of a triangle)
(b)
∠KLM = ∠KHM = 124° (Parallelogram)
∠JLM
= 124° - 14°
= 110°
∠p
= (180° - 110°) ÷ 2
= 35° (Isosceles triangle)
Answer(s): (a) 124°; (b) 35°