In the figure, PFGL is a trapezium and triangles PLK and KPH are isosceles triangles. NE, NG and EG are straight lines. PL = PK = KH. Find
- ∠p
- ∠q
(a)
∠FPH = 180° - ∠p (Interior angles, PF//NG)
∠KHP = 180° - ∠p (Angles on a straight line)
∠KPH = 180° - ∠p (Isosceles triangle)
53° + 180° - ∠p + 180° - ∠p + ∠q + 18° = 180° (Angles on a straight line, EN)
53° + 180° + 180° + 18° - ∠p - ∠p + ∠q = 180°
431° - 2∠p + ∠q = 180°
2∠p - ∠q = 431° - 180°
2∠p - ∠q = 251°
∠q = 2∠p - 251° --- (1)
∠PKL
= ∠PLK
= 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 - 251°
4∠p - 2∠p= 540° - 251°
2∠p = 289°
∠p
= 289° ÷ 2
= 144.5°
(b)
From (1)
∠q
= 2∠p - 251°
= 289° - 251°
= 38°
Answer(s): (a) 144.5°; (b) 38°