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
- ∠e
- ∠f
(a)
∠PXS = 180° - ∠e (Interior angles, XP//VR)
∠TSX = 180° - ∠e (Angles on a straight line)
∠TXS = 180° - ∠e (Isosceles triangle)
49° + 180° - ∠e + 180° - ∠e + ∠f + 22° = 180° (Angles on a straight line, NV)
49° + 180° + 180° + 22° - ∠e - ∠e + ∠f = 180°
431° - 2∠e + ∠f = 180°
2∠e - ∠f = 431° - 180°
2∠e - ∠f = 251°
∠f = 2∠e - 251° --- (1)
∠XTU
= ∠XUT
= 2 x (180° - ∠e)
= 360° - 2∠e (Exterior angle of a triangle)
∠f = 180° - (360° - 2∠e) - (360° - 2∠e)
∠f = 180° - 360° + 2∠e - 360° + 2∠e
∠f = 180° - 360° - 360° + 2∠e + 2∠e
∠f = 4∠e - 540° (Angles sum of triangle)
∠f = 4∠e - 540° --- (2)
(2) = (1)
4∠e - 540° = 2∠e - 251°
4∠e - 2∠e= 540° - 251°
2∠e = 289°
∠e
= 289° ÷ 2
= 144.5°
(b)
From (1)
∠f
= 2∠e - 251°
= 289° - 251°
= 38°
Answer(s): (a) 144.5°; (b) 38°