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
- ∠w
- ∠x
(a)
∠SZU = 180° - ∠w (Interior angles, ZS//YT)
∠VUZ = 180° - ∠w (Angles on a straight line)
∠VZU = 180° - ∠w (Isosceles triangle)
48° + 180° - ∠w + 180° - ∠w + ∠x + 15° = 180° (Angles on a straight line, RY)
48° + 180° + 180° + 15° - ∠w - ∠w + ∠x = 180°
423° - 2∠w + ∠x = 180°
2∠w - ∠x = 423° - 180°
2∠w - ∠x = 243°
∠x = 2∠w - 243° --- (1)
∠ZVX
= ∠ZXV
= 2 x (180° - ∠w)
= 360° - 2∠w (Exterior angle of a triangle)
∠x = 180° - (360° - 2∠w) - (360° - 2∠w)
∠x = 180° - 360° + 2∠w - 360° + 2∠w
∠x = 180° - 360° - 360° + 2∠w + 2∠w
∠x = 4∠w - 540° (Angles sum of triangle)
∠x = 4∠w - 540° --- (2)
(2) = (1)
4∠w - 540° = 2∠w - 243°
4∠w - 2∠w= 540° - 243°
2∠w = 297°
∠w
= 297° ÷ 2
= 148.5°
(b)
From (1)
∠x
= 2∠w - 243°
= 297° - 243°
= 54°
Answer(s): (a) 148.5°; (b) 54°