The figure is not drawn to scale. ZACD is a square and YZD is an equilateral triangle. ABC is an isosceles triangle. YCB is a straight line.
- Find ∠DYC.
- Find ∠ABC.
(a)
∠CDZ = 90°
∠YDZ = 60° (Equilateral triangle YDB)
∠YDC
= 90° + 60°
= 150°
∠DYC
= (180° - 150°) ÷ 2
= 30° ÷ 2
= 15° (Isosceles triangle)
(b)
∠YCA
= 90° - ∠YCD;
= 90° - 15°
= 75°
∠BCA
= 180° - ∠YCA
= 180° - 75°
= 105° (Angles on a straight line)
∠ABC
= (180° - ∠BCC) ÷ 2
= (180° - 105°) ÷ 2
= 75° ÷ 2
= 37.5° (Isosceles triangle)
Answer(s): (a) 15°; (b) 37.5°
