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
- ∠t
- ∠v
(a)
∠SZU = 180° - ∠t (Interior angles, ZS//YT)
∠VUZ = 180° - ∠t (Angles on a straight line)
∠VZU = 180° - ∠t (Isosceles triangle)
52° + 180° - ∠t + 180° - ∠t + ∠v + 22° = 180° (Angles on a straight line, RY)
52° + 180° + 180° + 22° - ∠t - ∠t + ∠v = 180°
434° - 2∠t + ∠v = 180°
2∠t - ∠v = 434° - 180°
2∠t - ∠v = 254°
∠v = 2∠t - 254° --- (1)
∠ZVX
= ∠ZXV
= 2 x (180° - ∠t)
= 360° - 2∠t (Exterior angle of a triangle)
∠v = 180° - (360° - 2∠t) - (360° - 2∠t)
∠v = 180° - 360° + 2∠t - 360° + 2∠t
∠v = 180° - 360° - 360° + 2∠t + 2∠t
∠v = 4∠t - 540° (Angles sum of triangle)
∠v = 4∠t - 540° --- (2)
(2) = (1)
4∠t - 540° = 2∠t - 254°
4∠t - 2∠t= 540° - 254°
2∠t = 286°
∠t
= 286° ÷ 2
= 143°
(b)
From (1)
∠v
= 2∠t - 254°
= 286° - 254°
= 32°
Answer(s): (a) 143°; (b) 32°