In the figure, not drawn to scale, CDEF is a square, EFG is an equilateral triangle, EGHJ is a rhombus, and DE = GE. Find
- ∠GHJ
- ∠CGD
(a)
∠FEG = 60°
∠DEG
= 90° - 60°
= 30°
EG = ED
∠EGD
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
∠GHJ
= 180° - 75°
= 105° (Interior angles)
(b)
∠EDG = ∠EGD = 75°
∠CDG
= 90° - 75°
= 15°
∠CGD
= 180° - 15° - 15°
= 150° (Isosceles triangle)
Answer(s): (a) 105°; (b) 150°