The figure is not drawn to scale. YZBC is a square and XYC is an equilateral triangle. ZAB is an isosceles triangle. XBA is a straight line.
- Find ∠CXB.
- Find ∠ZAB.
(a)
∠BCY = 90°
∠XCY = 60° (Equilateral triangle XCB)
∠XCB
= 90° + 60°
= 150°
∠CXB
= (180° - 150°) ÷ 2
= 30° ÷ 2
= 15° (Isosceles triangle)
(b)
∠XBZ
= 90° - ∠XBC;
= 90° - 15°
= 75°
∠ABZ
= 180° - ∠XBZ
= 180° - 75°
= 105° (Angles on a straight line)
∠ZAB
= (180° - ∠ABC) ÷ 2
= (180° - 105°) ÷ 2
= 75° ÷ 2
= 37.5° (Isosceles triangle)
Answer(s): (a) 15°; (b) 37.5°
