In the figure, not drawn to scale. CDEF is a square. EFG is an equilateral triangle and CE is a straight line. Find
- Two-thirds of ∠CHG
- Six 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°
Two-thirds of ∠CHG
=
23 x 75°
= 50°
(b)
∠GHE
= 180° - ∠CHG
= 180° - 75°
= 105° (Angles on a straight line)
∠GEH
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Six times of ∠GEH
= 6 x 15°
= 90°
Answer(s): (a) 50°; (b) 90°
