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
- ∠a
- ∠b
(a)
∠PXS = 180° - ∠a (Interior angles, XP//VR)
∠TSX = 180° - ∠a (Angles on a straight line)
∠TXS = 180° - ∠a (Isosceles triangle)
50° + 180° - ∠a + 180° - ∠a + ∠b + 23° = 180° (Angles on a straight line, NV)
50° + 180° + 180° + 23° - ∠a - ∠a + ∠b = 180°
433° - 2∠a + ∠b = 180°
2∠a - ∠b = 433° - 180°
2∠a - ∠b = 253°
∠b = 2∠a - 253° --- (1)
∠XTU
= ∠XUT
= 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 - 253°
4∠a - 2∠a= 540° - 253°
2∠a = 287°
∠a
= 287° ÷ 2
= 143.5°
(b)
From (1)
∠b
= 2∠a - 253°
= 287° - 253°
= 34°
Answer(s): (a) 143.5°; (b) 34°