In the figure, not drawn to scale. ABCD is a square. CDE is an equilateral triangle and AC is a straight line. Find
- One-fifth of ∠AFE
- Nine times of ∠ECF
(a)
∠DAC = 45° (Right angle)
∠FDC = 60° (Equilateral triangle)
∠ADF
= 90° - ∠FDC
= 90° - 60°
= 30°
∠AFE
= 30° + 45° (Exterior angle of a triangle)
= 75°
One-fifth of ∠AFE
=
15 x 75°
= 15°
(b)
∠EFC
= 180° - ∠AFE
= 180° - 75°
= 105° (Angles on a straight line)
∠ECF
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Nine times of ∠ECF
= 9 x 15°
= 135°
Answer(s): (a) 15°; (b) 135°