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
- ∠s
- ∠t
(a)
∠SZU = 180° - ∠s (Interior angles, ZS//YT)
∠VUZ = 180° - ∠s (Angles on a straight line)
∠VZU = 180° - ∠s (Isosceles triangle)
55° + 180° - ∠s + 180° - ∠s + ∠t + 20° = 180° (Angles on a straight line, RY)
55° + 180° + 180° + 20° - ∠s - ∠s + ∠t = 180°
435° - 2∠s + ∠t = 180°
2∠s - ∠t = 435° - 180°
2∠s - ∠t = 255°
∠t = 2∠s - 255° --- (1)
∠ZVX
= ∠ZXV
= 2 x (180° - ∠s)
= 360° - 2∠s (Exterior angle of a triangle)
∠t = 180° - (360° - 2∠s) - (360° - 2∠s)
∠t = 180° - 360° + 2∠s - 360° + 2∠s
∠t = 180° - 360° - 360° + 2∠s + 2∠s
∠t = 4∠s - 540° (Angles sum of triangle)
∠t = 4∠s - 540° --- (2)
(2) = (1)
4∠s - 540° = 2∠s - 255°
4∠s - 2∠s= 540° - 255°
2∠s = 285°
∠s
= 285° ÷ 2
= 142.5°
(b)
From (1)
∠t
= 2∠s - 255°
= 285° - 255°
= 30°
Answer(s): (a) 142.5°; (b) 30°