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
- ∠p
- ∠q
(a)
∠RYT = 180° - ∠p (Interior angles, YR//XS)
∠UTY = 180° - ∠p (Angles on a straight line)
∠UYT = 180° - ∠p (Isosceles triangle)
50° + 180° - ∠p + 180° - ∠p + ∠q + 16° = 180° (Angles on a straight line, PX)
50° + 180° + 180° + 16° - ∠p - ∠p + ∠q = 180°
426° - 2∠p + ∠q = 180°
2∠p - ∠q = 426° - 180°
2∠p - ∠q = 246°
∠q = 2∠p - 246° --- (1)
∠YUV
= ∠YVU
= 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 - 246°
4∠p - 2∠p= 540° - 246°
2∠p = 294°
∠p
= 294° ÷ 2
= 147°
(b)
From (1)
∠q
= 2∠p - 246°
= 294° - 246°
= 48°
Answer(s): (a) 147°; (b) 48°