In the figure, CDEF and DHJK are two rhombuses. If ∠HJK = 106° and ∠CFE = 91°, calculate
- ∠JHK
- ∠JED
- ∠KLE
(a)
∠JHK
= (180° - 106°) ÷ 2
= 37° (Isosceles triangle)
(b)
∠JED = ∠CFE = 91° (Corresponding angles)
(c)
∠KLE
= 91° - 37°
= 54° (Exterior angle of a triangle)
Answer(s): (a) 37°; (b) 91°; (c) 54°