In the figure, CDEF and DHJK are two rhombuses. If ∠HJK = 105° and ∠CFE = 97°, calculate
- ∠JHK
- ∠JED
- ∠KLE
(a)
∠JHK
= (180° - 105°) ÷ 2
= 37.5° (Isosceles triangle)
(b)
∠JED = ∠CFE = 97° (Corresponding angles)
(c)
∠KLE
= 97° - 37.5°
= 59.5° (Exterior angle of a triangle)
Answer(s): (a) 37.5°; (b) 97°; (c) 59.5°