The figure is not drawn to scale. DEGH is a square and CDH is an equilateral triangle. EFG is an isosceles triangle. CGF is a straight line.
- Find ∠HCG.
- Find ∠EFG.
(a)
∠GHD = 90°
∠CHD = 60° (Equilateral triangle CHB)
∠CHG
= 90° + 60°
= 150°
∠HCG
= (180° - 150°) ÷ 2
= 30° ÷ 2
= 15° (Isosceles triangle)
(b)
∠CGE
= 90° - ∠CGH;
= 90° - 15°
= 75°
∠FGE
= 180° - ∠CGE
= 180° - 75°
= 105° (Angles on a straight line)
∠EFG
= (180° - ∠FGC) ÷ 2
= (180° - 105°) ÷ 2
= 75° ÷ 2
= 37.5° (Isosceles triangle)
Answer(s): (a) 15°; (b) 37.5°
