In the figure, not drawn to scale. XYZA is a square. ZAB is an equilateral triangle and XZ is a straight line. Find
- Three-fifths of ∠XCB
- Eight times of ∠BZC
(a)
∠AXZ = 45° (Right angle)
∠CAZ = 60° (Equilateral triangle)
∠XAC
= 90° - ∠CAZ
= 90° - 60°
= 30°
∠XCB
= 30° + 45° (Exterior angle of a triangle)
= 75°
Three-fifths of ∠XCB
=
35 x 75°
= 45°
(b)
∠BCZ
= 180° - ∠XCB
= 180° - 75°
= 105° (Angles on a straight line)
∠BZC
= 180° - 60° - 105°
= 15° (Angles sum of triangle)
Eight times of ∠BZC
= 8 x 15°
= 120°
Answer(s): (a) 45°; (b) 120°
