In the figure, ZSTX is a trapezium and triangles ZXV and VZU are isosceles triangles. YR, YT and RT are straight lines. ZX = ZV = VU. Find
- ∠m
- ∠n
(a)
∠SZU = 180° - ∠m (Interior angles, ZS//YT)
∠VUZ = 180° - ∠m (Angles on a straight line)
∠VZU = 180° - ∠m (Isosceles triangle)
48° + 180° - ∠m + 180° - ∠m + ∠n + 24° = 180° (Angles on a straight line, RY)
48° + 180° + 180° + 24° - ∠m - ∠m + ∠n = 180°
432° - 2∠m + ∠n = 180°
2∠m - ∠n = 432° - 180°
2∠m - ∠n = 252°
∠n = 2∠m - 252° --- (1)
∠ZVX
= ∠ZXV
= 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 - 252°
4∠m - 2∠m= 540° - 252°
2∠m = 288°
∠m
= 288° ÷ 2
= 144°
(b)
From (1)
∠n
= 2∠m - 252°
= 288° - 252°
= 36°
Answer(s): (a) 144°; (b) 36°