The figure is not drawn to scale. CDFG is a square and BCG is an equilateral triangle. DEF is an isosceles triangle. BFE is a straight line.
- Find ∠GBF.
- Find ∠DEF.
(a)
∠FGC = 90°
∠BGC = 60° (Equilateral triangle BGB)
∠BGF
= 90° + 60°
= 150°
∠GBF
= (180° - 150°) ÷ 2
= 30° ÷ 2
= 15° (Isosceles triangle)
(b)
∠BFD
= 90° - ∠BFG;
= 90° - 15°
= 75°
∠EFD
= 180° - ∠BFD
= 180° - 75°
= 105° (Angles on a straight line)
∠DEF
= (180° - ∠EFC) ÷ 2
= (180° - 105°) ÷ 2
= 75° ÷ 2
= 37.5° (Isosceles triangle)
Answer(s): (a) 15°; (b) 37.5°
