In the figure, not drawn to scale, ABCD is a square, CDE is an equilateral triangle, CEFG is a rhombus, and BC = EC. Find
- ∠EFG
- ∠AEB
(a)
∠DCE = 60°
∠BCE
= 90° - 60°
= 30°
CE = CB
∠CEB
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
∠EFG
= 180° - 75°
= 105° (Interior angles)
(b)
∠CBE = ∠CEB = 75°
∠ABE
= 90° - 75°
= 15°
∠AEB
= 180° - 15° - 15°
= 150° (Isosceles triangle)
Answer(s): (a) 105°; (b) 150°