In the figure, not drawn to scale. BCDE is a square. DEF is an equilateral triangle and BD is a straight line. Find
- One-fifth of ∠BGF
- Thrice 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-fifth of ∠BGF
=
15 x 75°
= 15°
(b)
∠FGD
= 180° - ∠BGF
= 180° - 75°
= 105° (Angles on a straight line)
∠FDG
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Thrice of ∠FDG
= 3 x 15°
= 45°
Answer(s): (a) 15°; (b) 45°
