In the figure, CDEF and DHJK are two rhombuses. If ∠HJK = 120° and ∠CFE = 96°, calculate
- ∠JHK
- ∠JED
- ∠KLE
(a)
∠JHK
= (180° - 120°) ÷ 2
= 30° (Isosceles triangle)
(b)
∠JED = ∠CFE = 96° (Corresponding angles)
(c)
∠KLE
= 96° - 30°
= 66° (Exterior angle of a triangle)
Answer(s): (a) 30°; (b) 96°; (c) 66°