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 = 46° and ∠MJN = 33°. Find
- ∠FJH
- ∠JLK
(a)
∠FJN
= (180° - 46°) ÷ 2
= 67° (Isosceles triangle)
∠FJH
= 180° - 67°
= 113° (Angles on a straight line)
(b)
∠LJK
= 67° - 33°
= 34°
∠JLK
= (180° - 34°) ÷ 2
= 73 ° (Isosceles triangle)
Answer(s): (a) 113°; (b) 73°