In the figure, CDEF and DHJK are two rhombuses. If ∠HJK = 119° and ∠CFE = 100°, calculate
- ∠JHK
- ∠JED
- ∠KLE
(a)
∠JHK
= (180° - 119°) ÷ 2
= 30.5° (Isosceles triangle)
(b)
∠JED = ∠CFE = 100° (Corresponding angles)
(c)
∠KLE
= 100° - 30.5°
= 69.5° (Exterior angle of a triangle)
Answer(s): (a) 30.5°; (b) 100°; (c) 69.5°