In the figure, not drawn to scale. ZABC is a square. BCD is an equilateral triangle and ZB is a straight line. Find
- Four-fifths of ∠ZED
- Thrice of ∠DBE
(a)
∠CZB = 45° (Right angle)
∠ECB = 60° (Equilateral triangle)
∠ZCE
= 90° - ∠ECB
= 90° - 60°
= 30°
∠ZED
= 30° + 45° (Exterior angle of a triangle)
= 75°
Four-fifths of ∠ZED
=
45 x 75°
= 60°
(b)
∠DEB
= 180° - ∠ZED
= 180° - 75°
= 105° (Angles on a straight line)
∠DBE
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Thrice of ∠DBE
= 3 x 15°
= 45°
Answer(s): (a) 60°; (b) 45°
