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
- ∠f
- ∠g
(a)
∠FPH = 180° - ∠f (Interior angles, PF//NG)
∠KHP = 180° - ∠f (Angles on a straight line)
∠KPH = 180° - ∠f (Isosceles triangle)
52° + 180° - ∠f + 180° - ∠f + ∠g + 17° = 180° (Angles on a straight line, EN)
52° + 180° + 180° + 17° - ∠f - ∠f + ∠g = 180°
429° - 2∠f + ∠g = 180°
2∠f - ∠g = 429° - 180°
2∠f - ∠g = 249°
∠g = 2∠f - 249° --- (1)
∠PKL
= ∠PLK
= 2 x (180° - ∠f)
= 360° - 2∠f (Exterior angle of a triangle)
∠g = 180° - (360° - 2∠f) - (360° - 2∠f)
∠g = 180° - 360° + 2∠f - 360° + 2∠f
∠g = 180° - 360° - 360° + 2∠f + 2∠f
∠g = 4∠f - 540° (Angles sum of triangle)
∠g = 4∠f - 540° --- (2)
(2) = (1)
4∠f - 540° = 2∠f - 249°
4∠f - 2∠f= 540° - 249°
2∠f = 291°
∠f
= 291° ÷ 2
= 145.5°
(b)
From (1)
∠g
= 2∠f - 249°
= 291° - 249°
= 42°
Answer(s): (a) 145.5°; (b) 42°