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
- ∠a
- ∠b
(a)
∠SZU = 180° - ∠a (Interior angles, ZS//YT)
∠VUZ = 180° - ∠a (Angles on a straight line)
∠VZU = 180° - ∠a (Isosceles triangle)
58° + 180° - ∠a + 180° - ∠a + ∠b + 16° = 180° (Angles on a straight line, RY)
58° + 180° + 180° + 16° - ∠a - ∠a + ∠b = 180°
434° - 2∠a + ∠b = 180°
2∠a - ∠b = 434° - 180°
2∠a - ∠b = 254°
∠b = 2∠a - 254° --- (1)
∠ZVX
= ∠ZXV
= 2 x (180° - ∠a)
= 360° - 2∠a (Exterior angle of a triangle)
∠b = 180° - (360° - 2∠a) - (360° - 2∠a)
∠b = 180° - 360° + 2∠a - 360° + 2∠a
∠b = 180° - 360° - 360° + 2∠a + 2∠a
∠b = 4∠a - 540° (Angles sum of triangle)
∠b = 4∠a - 540° --- (2)
(2) = (1)
4∠a - 540° = 2∠a - 254°
4∠a - 2∠a= 540° - 254°
2∠a = 286°
∠a
= 286° ÷ 2
= 143°
(b)
From (1)
∠b
= 2∠a - 254°
= 286° - 254°
= 32°
Answer(s): (a) 143°; (b) 32°