In the figure, not drawn to scale. WXYZ is a square. YZA is an equilateral triangle and WY is a straight line. Find
- Four-fifths of ∠WBA
- Six times of ∠AYB
(a)
∠ZWY = 45° (Right angle)
∠BZY = 60° (Equilateral triangle)
∠WZB
= 90° - ∠BZY
= 90° - 60°
= 30°
∠WBA
= 30° + 45° (Exterior angle of a triangle)
= 75°
Four-fifths of ∠WBA
=
45 x 75°
= 60°
(b)
∠ABY
= 180° - ∠WBA
= 180° - 75°
= 105° (Angles on a straight line)
∠AYB
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Six times of ∠AYB
= 6 x 15°
= 90°
Answer(s): (a) 60°; (b) 90°
