In the figure, FGHJ is a trapezium and FJKN is a rhombus. FJ is parallel to GH and JK = JL. NJH and JML are straight lines. ∠JFN = 40° and ∠MJN = 28°. Find
- ∠FJH
- ∠JLK
(a)
∠FJN
= (180° - 40°) ÷ 2
= 70° (Isosceles triangle)
∠FJH
= 180° - 70°
= 110° (Angles on a straight line)
(b)
∠LJK
= 70° - 28°
= 42°
∠JLK
= (180° - 42°) ÷ 2
= 69 ° (Isosceles triangle)
Answer(s): (a) 110°; (b) 69°