The figure is not drawn to scale. YZC is an equilateral triangle and BZA is an isosceles triangle. AZY and ZBD are straight lines and AY//DC. ∠CZD = 90° and ∠BCZ = 58°.
- Find ∠AZD.
- Find ∠ZBA.
- Find ∠CBD.
(a)
∠CZY = 60° (Equilateral triangle YZY)
∠CZD = 90°
∠BCZ = 58°
∠AZD
= 180° - ∠CZD - ∠CZY
= 180° - 90° - 60°
= 30° (Angles on a straight line)
(b)
∠ZBA
= (180° - ∠AZD ÷ 2
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
(c)
∠CBZ
= 180° - ∠CZD - ∠BCZ
= 180° - 90° - 58°
= 32°
∠CBD
= 180° - ∠CBZ
= 180° - 32°
= 148° (Angles on a straight line)
Answer(s): (a) 30°; (b) 75°; (c) 148°
