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
- ∠p
- ∠q
(a)
∠PXS = 180° - ∠p (Interior angles, XP//VR)
∠TSX = 180° - ∠p (Angles on a straight line)
∠TXS = 180° - ∠p (Isosceles triangle)
52° + 180° - ∠p + 180° - ∠p + ∠q + 18° = 180° (Angles on a straight line, NV)
52° + 180° + 180° + 18° - ∠p - ∠p + ∠q = 180°
430° - 2∠p + ∠q = 180°
2∠p - ∠q = 430° - 180°
2∠p - ∠q = 250°
∠q = 2∠p - 250° --- (1)
∠XTU
= ∠XUT
= 2 x (180° - ∠p)
= 360° - 2∠p (Exterior angle of a triangle)
∠q = 180° - (360° - 2∠p) - (360° - 2∠p)
∠q = 180° - 360° + 2∠p - 360° + 2∠p
∠q = 180° - 360° - 360° + 2∠p + 2∠p
∠q = 4∠p - 540° (Angles sum of triangle)
∠q = 4∠p - 540° --- (2)
(2) = (1)
4∠p - 540° = 2∠p - 250°
4∠p - 2∠p= 540° - 250°
2∠p = 290°
∠p
= 290° ÷ 2
= 145°
(b)
From (1)
∠q
= 2∠p - 250°
= 290° - 250°
= 40°
Answer(s): (a) 145°; (b) 40°