The figure is not drawn to scale. Triangle CDB is an isosceles triangle. Triangle ZAB is an equilateral triangle. ∠DBC is
25 of ∠AZB and ∠ABD = ∠ZBC.
- Find ∠ABD.
- Find ∠ZCB.
(a)
∠ABZ = 60° (Equilateral triangle)
∠DBC
= ∠ABZ x
25 = 60° x
25 = 24°
∠ABD = ∠ZBC
∠ABD
= ∠ABZ - ∠DBA) ÷ 2
= (60° - 24°) ÷ 2
= 36° ÷ 2
= 18° (Isosceles triangle)
(b)
∠ZCB
= 180° - ∠AZB - ∠ABD
= 180° - 60° - 18°
= 102° (Angles sum of triangle)
Answer(s): (a) 18°; (b) 102°