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 = 48° and ∠MJN = 27°. Find
- ∠FJH
- ∠JLK
(a)
∠FJN
= (180° - 48°) ÷ 2
= 66° (Isosceles triangle)
∠FJH
= 180° - 66°
= 114° (Angles on a straight line)
(b)
∠LJK
= 66° - 27°
= 39°
∠JLK
= (180° - 39°) ÷ 2
= 70.5 ° (Isosceles triangle)
Answer(s): (a) 114°; (b) 70.5°