In the figure, XPRU is a trapezium and triangles XUT and TXS are isosceles triangles. VN, VR and NR are straight lines. XU = XT = TS. Find
- ∠q
- ∠r
(a)
∠PXS = 180° - ∠q (Interior angles, XP//VR)
∠TSX = 180° - ∠q (Angles on a straight line)
∠TXS = 180° - ∠q (Isosceles triangle)
51° + 180° - ∠q + 180° - ∠q + ∠r + 16° = 180° (Angles on a straight line, NV)
51° + 180° + 180° + 16° - ∠q - ∠q + ∠r = 180°
427° - 2∠q + ∠r = 180°
2∠q - ∠r = 427° - 180°
2∠q - ∠r = 247°
∠r = 2∠q - 247° --- (1)
∠XTU
= ∠XUT
= 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 - 247°
4∠q - 2∠q= 540° - 247°
2∠q = 293°
∠q
= 293° ÷ 2
= 146.5°
(b)
From (1)
∠r
= 2∠q - 247°
= 293° - 247°
= 46°
Answer(s): (a) 146.5°; (b) 46°