PSLEIn the figure, EFGH is a straight line, FHJK is a parallelogram and JG = JK. ∠KEF is a right angle, ∠EKF = 36° and ∠HJG = 14°.
- Find ∠l.
- Find ∠m.
(a)
∠KEF = 90° (Right angle)
∠l
= 90° + 36°
= 126° (Exterior angle of a triangle)
(b)
∠HJK = ∠HFK = 126° (Parallelogram)
∠GJK
= 126° - 14°
= 112°
∠m
= (180° - 112°) ÷ 2
= 34° (Isosceles triangle)
Answer(s): (a) 126°; (b) 34°