In the figure, LDEH is a trapezium and triangles LHG and GLF are isosceles triangles. KC, KE and CE are straight lines. LH = LG = GF. Find
- ∠g
- ∠h
(a)
∠DLF = 180° - ∠g (Interior angles, LD//KE)
∠GFL = 180° - ∠g (Angles on a straight line)
∠GLF = 180° - ∠g (Isosceles triangle)
53° + 180° - ∠g + 180° - ∠g + ∠h + 23° = 180° (Angles on a straight line, CK)
53° + 180° + 180° + 23° - ∠g - ∠g + ∠h = 180°
436° - 2∠g + ∠h = 180°
2∠g - ∠h = 436° - 180°
2∠g - ∠h = 256°
∠h = 2∠g - 256° --- (1)
∠LGH
= ∠LHG
= 2 x (180° - ∠g)
= 360° - 2∠g (Exterior angle of a triangle)
∠h = 180° - (360° - 2∠g) - (360° - 2∠g)
∠h = 180° - 360° + 2∠g - 360° + 2∠g
∠h = 180° - 360° - 360° + 2∠g + 2∠g
∠h = 4∠g - 540° (Angles sum of triangle)
∠h = 4∠g - 540° --- (2)
(2) = (1)
4∠g - 540° = 2∠g - 256°
4∠g - 2∠g= 540° - 256°
2∠g = 284°
∠g
= 284° ÷ 2
= 142°
(b)
From (1)
∠h
= 2∠g - 256°
= 284° - 256°
= 28°
Answer(s): (a) 142°; (b) 28°