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
- ∠q
- ∠r
(a)
∠SZU = 180° - ∠q (Interior angles, ZS//YT)
∠VUZ = 180° - ∠q (Angles on a straight line)
∠VZU = 180° - ∠q (Isosceles triangle)
57° + 180° - ∠q + 180° - ∠q + ∠r + 25° = 180° (Angles on a straight line, RY)
57° + 180° + 180° + 25° - ∠q - ∠q + ∠r = 180°
442° - 2∠q + ∠r = 180°
2∠q - ∠r = 442° - 180°
2∠q - ∠r = 262°
∠r = 2∠q - 262° --- (1)
∠ZVX
= ∠ZXV
= 2 x (180° - ∠q)
= 360° - 2∠q (Exterior angle of a triangle)
∠r = 180° - (360° - 2∠q) - (360° - 2∠q)
∠r = 180° - 360° + 2∠q - 360° + 2∠q
∠r = 180° - 360° - 360° + 2∠q + 2∠q
∠r = 4∠q - 540° (Angles sum of triangle)
∠r = 4∠q - 540° --- (2)
(2) = (1)
4∠q - 540° = 2∠q - 262°
4∠q - 2∠q= 540° - 262°
2∠q = 278°
∠q
= 278° ÷ 2
= 139°
(b)
From (1)
∠r
= 2∠q - 262°
= 278° - 262°
= 16°
Answer(s): (a) 139°; (b) 16°