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
- ∠v
- ∠w
(a)
∠PXS = 180° - ∠v (Interior angles, XP//VR)
∠TSX = 180° - ∠v (Angles on a straight line)
∠TXS = 180° - ∠v (Isosceles triangle)
54° + 180° - ∠v + 180° - ∠v + ∠w + 24° = 180° (Angles on a straight line, NV)
54° + 180° + 180° + 24° - ∠v - ∠v + ∠w = 180°
438° - 2∠v + ∠w = 180°
2∠v - ∠w = 438° - 180°
2∠v - ∠w = 258°
∠w = 2∠v - 258° --- (1)
∠XTU
= ∠XUT
= 2 x (180° - ∠v)
= 360° - 2∠v (Exterior angle of a triangle)
∠w = 180° - (360° - 2∠v) - (360° - 2∠v)
∠w = 180° - 360° + 2∠v - 360° + 2∠v
∠w = 180° - 360° - 360° + 2∠v + 2∠v
∠w = 4∠v - 540° (Angles sum of triangle)
∠w = 4∠v - 540° --- (2)
(2) = (1)
4∠v - 540° = 2∠v - 258°
4∠v - 2∠v= 540° - 258°
2∠v = 282°
∠v
= 282° ÷ 2
= 141°
(b)
From (1)
∠w
= 2∠v - 258°
= 282° - 258°
= 24°
Answer(s): (a) 141°; (b) 24°