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
- ∠b
- ∠c
(a)
∠PXS = 180° - ∠b (Interior angles, XP//VR)
∠TSX = 180° - ∠b (Angles on a straight line)
∠TXS = 180° - ∠b (Isosceles triangle)
49° + 180° - ∠b + 180° - ∠b + ∠c + 24° = 180° (Angles on a straight line, NV)
49° + 180° + 180° + 24° - ∠b - ∠b + ∠c = 180°
433° - 2∠b + ∠c = 180°
2∠b - ∠c = 433° - 180°
2∠b - ∠c = 253°
∠c = 2∠b - 253° --- (1)
∠XTU
= ∠XUT
= 2 x (180° - ∠b)
= 360° - 2∠b (Exterior angle of a triangle)
∠c = 180° - (360° - 2∠b) - (360° - 2∠b)
∠c = 180° - 360° + 2∠b - 360° + 2∠b
∠c = 180° - 360° - 360° + 2∠b + 2∠b
∠c = 4∠b - 540° (Angles sum of triangle)
∠c = 4∠b - 540° --- (2)
(2) = (1)
4∠b - 540° = 2∠b - 253°
4∠b - 2∠b= 540° - 253°
2∠b = 287°
∠b
= 287° ÷ 2
= 143.5°
(b)
From (1)
∠c
= 2∠b - 253°
= 287° - 253°
= 34°
Answer(s): (a) 143.5°; (b) 34°