In the figure, not drawn to scale. CDEF is a square. EFG is an equilateral triangle and CE is a straight line. Find
- Three-fifths of ∠CHG
- Nine times of ∠GEH
(a)
∠FCE = 45° (Right angle)
∠HFE = 60° (Equilateral triangle)
∠CFH
= 90° - ∠HFE
= 90° - 60°
= 30°
∠CHG
= 30° + 45° (Exterior angle of a triangle)
= 75°
Three-fifths of ∠CHG
=
35 x 75°
= 45°
(b)
∠GHE
= 180° - ∠CHG
= 180° - 75°
= 105° (Angles on a straight line)
∠GEH
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Nine times of ∠GEH
= 9 x 15°
= 135°
Answer(s): (a) 45°; (b) 135°
