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
- ∠i
- ∠j
(a)
∠SZU = 180° - ∠i (Interior angles, ZS//YT)
∠VUZ = 180° - ∠i (Angles on a straight line)
∠VZU = 180° - ∠i (Isosceles triangle)
55° + 180° - ∠i + 180° - ∠i + ∠j + 24° = 180° (Angles on a straight line, RY)
55° + 180° + 180° + 24° - ∠i - ∠i + ∠j = 180°
439° - 2∠i + ∠j = 180°
2∠i - ∠j = 439° - 180°
2∠i - ∠j = 259°
∠j = 2∠i - 259° --- (1)
∠ZVX
= ∠ZXV
= 2 x (180° - ∠i)
= 360° - 2∠i (Exterior angle of a triangle)
∠j = 180° - (360° - 2∠i) - (360° - 2∠i)
∠j = 180° - 360° + 2∠i - 360° + 2∠i
∠j = 180° - 360° - 360° + 2∠i + 2∠i
∠j = 4∠i - 540° (Angles sum of triangle)
∠j = 4∠i - 540° --- (2)
(2) = (1)
4∠i - 540° = 2∠i - 259°
4∠i - 2∠i= 540° - 259°
2∠i = 281°
∠i
= 281° ÷ 2
= 140.5°
(b)
From (1)
∠j
= 2∠i - 259°
= 281° - 259°
= 22°
Answer(s): (a) 140.5°; (b) 22°