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
- ∠e
- ∠f
(a)
∠DLF = 180° - ∠e (Interior angles, LD//KE)
∠GFL = 180° - ∠e (Angles on a straight line)
∠GLF = 180° - ∠e (Isosceles triangle)
57° + 180° - ∠e + 180° - ∠e + ∠f + 19° = 180° (Angles on a straight line, CK)
57° + 180° + 180° + 19° - ∠e - ∠e + ∠f = 180°
436° - 2∠e + ∠f = 180°
2∠e - ∠f = 436° - 180°
2∠e - ∠f = 256°
∠f = 2∠e - 256° --- (1)
∠LGH
= ∠LHG
= 2 x (180° - ∠e)
= 360° - 2∠e (Exterior angle of a triangle)
∠f = 180° - (360° - 2∠e) - (360° - 2∠e)
∠f = 180° - 360° + 2∠e - 360° + 2∠e
∠f = 180° - 360° - 360° + 2∠e + 2∠e
∠f = 4∠e - 540° (Angles sum of triangle)
∠f = 4∠e - 540° --- (2)
(2) = (1)
4∠e - 540° = 2∠e - 256°
4∠e - 2∠e= 540° - 256°
2∠e = 284°
∠e
= 284° ÷ 2
= 142°
(b)
From (1)
∠f
= 2∠e - 256°
= 284° - 256°
= 28°
Answer(s): (a) 142°; (b) 28°