In the figure, CDEF and DHJK are two rhombuses. If ∠HJK = 113° and ∠CFE = 92°, calculate
- ∠JHK
- ∠JED
- ∠KLE
(a)
∠JHK
= (180° - 113°) ÷ 2
= 33.5° (Isosceles triangle)
(b)
∠JED = ∠CFE = 92° (Corresponding angles)
(c)
∠KLE
= 92° - 33.5°
= 58.5° (Exterior angle of a triangle)
Answer(s): (a) 33.5°; (b) 92°; (c) 58.5°