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 = 45° and ∠MJN = 35°. Find
- ∠FJH
- ∠JLK
(a)
∠FJN
= (180° - 45°) ÷ 2
= 67.5° (Isosceles triangle)
∠FJH
= 180° - 67.5°
= 112.5° (Angles on a straight line)
(b)
∠LJK
= 67.5° - 35°
= 32.5°
∠JLK
= (180° - 32.5°) ÷ 2
= 73.75 ° (Isosceles triangle)
Answer(s): (a) 112.5°; (b) 73.75°