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
- ∠m
- ∠n
(a)
∠PXS = 180° - ∠m (Interior angles, XP//VR)
∠TSX = 180° - ∠m (Angles on a straight line)
∠TXS = 180° - ∠m (Isosceles triangle)
58° + 180° - ∠m + 180° - ∠m + ∠n + 24° = 180° (Angles on a straight line, NV)
58° + 180° + 180° + 24° - ∠m - ∠m + ∠n = 180°
442° - 2∠m + ∠n = 180°
2∠m - ∠n = 442° - 180°
2∠m - ∠n = 262°
∠n = 2∠m - 262° --- (1)
∠XTU
= ∠XUT
= 2 x (180° - ∠m)
= 360° - 2∠m (Exterior angle of a triangle)
∠n = 180° - (360° - 2∠m) - (360° - 2∠m)
∠n = 180° - 360° + 2∠m - 360° + 2∠m
∠n = 180° - 360° - 360° + 2∠m + 2∠m
∠n = 4∠m - 540° (Angles sum of triangle)
∠n = 4∠m - 540° --- (2)
(2) = (1)
4∠m - 540° = 2∠m - 262°
4∠m - 2∠m= 540° - 262°
2∠m = 278°
∠m
= 278° ÷ 2
= 139°
(b)
From (1)
∠n
= 2∠m - 262°
= 278° - 262°
= 16°
Answer(s): (a) 139°; (b) 16°