In the figure, ZSTX is a trapezium and triangles ZXV and VZU are isosceles triangles. YR, YT and RT are straight lines. ZX = ZV = VU. Find
- ∠a
- ∠b
(a)
∠SZU = 180° - ∠a (Interior angles, ZS//YT)
∠VUZ = 180° - ∠a (Angles on a straight line)
∠VZU = 180° - ∠a (Isosceles triangle)
53° + 180° - ∠a + 180° - ∠a + ∠b + 22° = 180° (Angles on a straight line, RY)
53° + 180° + 180° + 22° - ∠a - ∠a + ∠b = 180°
435° - 2∠a + ∠b = 180°
2∠a - ∠b = 435° - 180°
2∠a - ∠b = 255°
∠b = 2∠a - 255° --- (1)
∠ZVX
= ∠ZXV
= 2 x (180° - ∠a)
= 360° - 2∠a (Exterior angle of a triangle)
∠b = 180° - (360° - 2∠a) - (360° - 2∠a)
∠b = 180° - 360° + 2∠a - 360° + 2∠a
∠b = 180° - 360° - 360° + 2∠a + 2∠a
∠b = 4∠a - 540° (Angles sum of triangle)
∠b = 4∠a - 540° --- (2)
(2) = (1)
4∠a - 540° = 2∠a - 255°
4∠a - 2∠a= 540° - 255°
2∠a = 285°
∠a
= 285° ÷ 2
= 142.5°
(b)
From (1)
∠b
= 2∠a - 255°
= 285° - 255°
= 30°
Answer(s): (a) 142.5°; (b) 30°