The figure is not drawn to scale. Triangle CDB is an isosceles triangle. Triangle ZAB is an equilateral triangle. ∠DBC is
35 of ∠AZB and ∠ABD = ∠ZBC.
- Find ∠ABD.
- Find ∠ZCB.
(a)
∠ABZ = 60° (Equilateral triangle)
∠DBC
= ∠ABZ x
35 = 60° x
35 = 36°
∠ABD = ∠ZBC
∠ABD
= ∠ABZ - ∠DBA) ÷ 2
= (60° - 36°) ÷ 2
= 24° ÷ 2
= 12° (Isosceles triangle)
(b)
∠ZCB
= 180° - ∠AZB - ∠ABD
= 180° - 60° - 12°
= 108° (Angles sum of triangle)
Answer(s): (a) 12°; (b) 108°