In the figure, CDEF and DHJK are two rhombuses. If ∠HJK = 108° and ∠CFE = 93°, calculate
- ∠JHK
- ∠JED
- ∠KLE
(a)
∠JHK
= (180° - 108°) ÷ 2
= 36° (Isosceles triangle)
(b)
∠JED = ∠CFE = 93° (Corresponding angles)
(c)
∠KLE
= 93° - 36°
= 57° (Exterior angle of a triangle)
Answer(s): (a) 36°; (b) 93°; (c) 57°