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
- ∠g
- ∠h
(a)
∠SZU = 180° - ∠g (Interior angles, ZS//YT)
∠VUZ = 180° - ∠g (Angles on a straight line)
∠VZU = 180° - ∠g (Isosceles triangle)
49° + 180° - ∠g + 180° - ∠g + ∠h + 24° = 180° (Angles on a straight line, RY)
49° + 180° + 180° + 24° - ∠g - ∠g + ∠h = 180°
433° - 2∠g + ∠h = 180°
2∠g - ∠h = 433° - 180°
2∠g - ∠h = 253°
∠h = 2∠g - 253° --- (1)
∠ZVX
= ∠ZXV
= 2 x (180° - ∠g)
= 360° - 2∠g (Exterior angle of a triangle)
∠h = 180° - (360° - 2∠g) - (360° - 2∠g)
∠h = 180° - 360° + 2∠g - 360° + 2∠g
∠h = 180° - 360° - 360° + 2∠g + 2∠g
∠h = 4∠g - 540° (Angles sum of triangle)
∠h = 4∠g - 540° --- (2)
(2) = (1)
4∠g - 540° = 2∠g - 253°
4∠g - 2∠g= 540° - 253°
2∠g = 287°
∠g
= 287° ÷ 2
= 143.5°
(b)
From (1)
∠h
= 2∠g - 253°
= 287° - 253°
= 34°
Answer(s): (a) 143.5°; (b) 34°