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 = 49° and ∠MJN = 26°. Find
- ∠FJH
- ∠JLK
(a)
∠FJN
= (180° - 49°) ÷ 2
= 65.5° (Isosceles triangle)
∠FJH
= 180° - 65.5°
= 114.5° (Angles on a straight line)
(b)
∠LJK
= 65.5° - 26°
= 39.5°
∠JLK
= (180° - 39.5°) ÷ 2
= 70.25 ° (Isosceles triangle)
Answer(s): (a) 114.5°; (b) 70.25°