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
- ∠r
- ∠s
(a)
∠SZU = 180° - ∠r (Interior angles, ZS//YT)
∠VUZ = 180° - ∠r (Angles on a straight line)
∠VZU = 180° - ∠r (Isosceles triangle)
54° + 180° - ∠r + 180° - ∠r + ∠s + 20° = 180° (Angles on a straight line, RY)
54° + 180° + 180° + 20° - ∠r - ∠r + ∠s = 180°
434° - 2∠r + ∠s = 180°
2∠r - ∠s = 434° - 180°
2∠r - ∠s = 254°
∠s = 2∠r - 254° --- (1)
∠ZVX
= ∠ZXV
= 2 x (180° - ∠r)
= 360° - 2∠r (Exterior angle of a triangle)
∠s = 180° - (360° - 2∠r) - (360° - 2∠r)
∠s = 180° - 360° + 2∠r - 360° + 2∠r
∠s = 180° - 360° - 360° + 2∠r + 2∠r
∠s = 4∠r - 540° (Angles sum of triangle)
∠s = 4∠r - 540° --- (2)
(2) = (1)
4∠r - 540° = 2∠r - 254°
4∠r - 2∠r= 540° - 254°
2∠r = 286°
∠r
= 286° ÷ 2
= 143°
(b)
From (1)
∠s
= 2∠r - 254°
= 286° - 254°
= 32°
Answer(s): (a) 143°; (b) 32°