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
- ∠r
- ∠s
(a)
∠PXS = 180° - ∠r (Interior angles, XP//VR)
∠TSX = 180° - ∠r (Angles on a straight line)
∠TXS = 180° - ∠r (Isosceles triangle)
50° + 180° - ∠r + 180° - ∠r + ∠s + 16° = 180° (Angles on a straight line, NV)
50° + 180° + 180° + 16° - ∠r - ∠r + ∠s = 180°
426° - 2∠r + ∠s = 180°
2∠r - ∠s = 426° - 180°
2∠r - ∠s = 246°
∠s = 2∠r - 246° --- (1)
∠XTU
= ∠XUT
= 2 x (180° - ∠r)
= 360° - 2∠r (Exterior angle of a triangle)
∠s = 180° - (360° - 2∠r) - (360° - 2∠r)
∠s = 180° - 360° + 2∠r - 360° + 2∠r
∠s = 180° - 360° - 360° + 2∠r + 2∠r
∠s = 4∠r - 540° (Angles sum of triangle)
∠s = 4∠r - 540° --- (2)
(2) = (1)
4∠r - 540° = 2∠r - 246°
4∠r - 2∠r= 540° - 246°
2∠r = 294°
∠r
= 294° ÷ 2
= 147°
(b)
From (1)
∠s
= 2∠r - 246°
= 294° - 246°
= 48°
Answer(s): (a) 147°; (b) 48°