In the figure, not drawn to scale. DEFG is a square. FGH is an equilateral triangle and DF is a straight line. Find
- One-fifth of ∠DJH
- Four times of ∠HFJ
(a)
∠GDF = 45° (Right angle)
∠JGF = 60° (Equilateral triangle)
∠DGJ
= 90° - ∠JGF
= 90° - 60°
= 30°
∠DJH
= 30° + 45° (Exterior angle of a triangle)
= 75°
One-fifth of ∠DJH
=
15 x 75°
= 15°
(b)
∠HJF
= 180° - ∠DJH
= 180° - 75°
= 105° (Angles on a straight line)
∠HFJ
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Four times of ∠HFJ
= 4 x 15°
= 60°
Answer(s): (a) 15°; (b) 60°
