In the figure, not drawn to scale. BCDE is a square. DEF is an equilateral triangle and BD is a straight line. Find
- One-third of ∠BGF
- Six times of ∠FDG
(a)
∠EBD = 45° (Right angle)
∠GED = 60° (Equilateral triangle)
∠BEG
= 90° - ∠GED
= 90° - 60°
= 30°
∠BGF
= 30° + 45° (Exterior angle of a triangle)
= 75°
One-third of ∠BGF
=
13 x 75°
= 25°
(b)
∠FGD
= 180° - ∠BGF
= 180° - 75°
= 105° (Angles on a straight line)
∠FDG
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Six times of ∠FDG
= 6 x 15°
= 90°
Answer(s): (a) 25°; (b) 90°