In the figure, YRSV is a trapezium and triangles YVU and UYT are isosceles triangles. XP, XS and PS are straight lines. YV = YU = UT. Find
- ∠n
- ∠p
(a)
∠RYT = 180° - ∠n (Interior angles, YR//XS)
∠UTY = 180° - ∠n (Angles on a straight line)
∠UYT = 180° - ∠n (Isosceles triangle)
55° + 180° - ∠n + 180° - ∠n + ∠p + 23° = 180° (Angles on a straight line, PX)
55° + 180° + 180° + 23° - ∠n - ∠n + ∠p = 180°
438° - 2∠n + ∠p = 180°
2∠n - ∠p = 438° - 180°
2∠n - ∠p = 258°
∠p = 2∠n - 258° --- (1)
∠YUV
= ∠YVU
= 2 x (180° - ∠n)
= 360° - 2∠n (Exterior angle of a triangle)
∠p = 180° - (360° - 2∠n) - (360° - 2∠n)
∠p = 180° - 360° + 2∠n - 360° + 2∠n
∠p = 180° - 360° - 360° + 2∠n + 2∠n
∠p = 4∠n - 540° (Angles sum of triangle)
∠p = 4∠n - 540° --- (2)
(2) = (1)
4∠n - 540° = 2∠n - 258°
4∠n - 2∠n= 540° - 258°
2∠n = 282°
∠n
= 282° ÷ 2
= 141°
(b)
From (1)
∠p
= 2∠n - 258°
= 282° - 258°
= 24°
Answer(s): (a) 141°; (b) 24°