In the figure, CDEF and DHJK are two rhombuses. If ∠HJK = 118° and ∠CFE = 95°, calculate
- ∠JHK
- ∠JED
- ∠KLE
(a)
∠JHK
= (180° - 118°) ÷ 2
= 31° (Isosceles triangle)
(b)
∠JED = ∠CFE = 95° (Corresponding angles)
(c)
∠KLE
= 95° - 31°
= 64° (Exterior angle of a triangle)
Answer(s): (a) 31°; (b) 95°; (c) 64°