In the figure, HBCF is a trapezium and triangles HFE and EHD are isosceles triangles. GA, GC and AC are straight lines. HF = HE = ED. Find
- ∠e
- ∠f
(a)
∠BHD = 180° - ∠e (Interior angles, HB//GC)
∠EDH = 180° - ∠e (Angles on a straight line)
∠EHD = 180° - ∠e (Isosceles triangle)
52° + 180° - ∠e + 180° - ∠e + ∠f + 16° = 180° (Angles on a straight line, AG)
52° + 180° + 180° + 16° - ∠e - ∠e + ∠f = 180°
428° - 2∠e + ∠f = 180°
2∠e - ∠f = 428° - 180°
2∠e - ∠f = 248°
∠f = 2∠e - 248° --- (1)
∠HEF
= ∠HFE
= 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 - 248°
4∠e - 2∠e= 540° - 248°
2∠e = 292°
∠e
= 292° ÷ 2
= 146°
(b)
From (1)
∠f
= 2∠e - 248°
= 292° - 248°
= 44°
Answer(s): (a) 146°; (b) 44°