In the figure, CDEF and DHJK are two rhombuses. If ∠HJK = 124° and ∠CFE = 102°, calculate
- ∠JHK
- ∠JED
- ∠KLE
(a)
∠JHK
= (180° - 124°) ÷ 2
= 28° (Isosceles triangle)
(b)
∠JED = ∠CFE = 102° (Corresponding angles)
(c)
∠KLE
= 102° - 28°
= 74° (Exterior angle of a triangle)
Answer(s): (a) 28°; (b) 102°; (c) 74°