In the figure, KCDG is a trapezium and triangles KGF and FKE are isosceles triangles. HB, HD and BD are straight lines. KG = KF = FE. Find
- ∠c
- ∠d
(a)
∠CKE = 180° - ∠c (Interior angles, KC//HD)
∠FEK = 180° - ∠c (Angles on a straight line)
∠FKE = 180° - ∠c (Isosceles triangle)
56° + 180° - ∠c + 180° - ∠c + ∠d + 25° = 180° (Angles on a straight line, BH)
56° + 180° + 180° + 25° - ∠c - ∠c + ∠d = 180°
441° - 2∠c + ∠d = 180°
2∠c - ∠d = 441° - 180°
2∠c - ∠d = 261°
∠d = 2∠c - 261° --- (1)
∠KFG
= ∠KGF
= 2 x (180° - ∠c)
= 360° - 2∠c (Exterior angle of a triangle)
∠d = 180° - (360° - 2∠c) - (360° - 2∠c)
∠d = 180° - 360° + 2∠c - 360° + 2∠c
∠d = 180° - 360° - 360° + 2∠c + 2∠c
∠d = 4∠c - 540° (Angles sum of triangle)
∠d = 4∠c - 540° --- (2)
(2) = (1)
4∠c - 540° = 2∠c - 261°
4∠c - 2∠c= 540° - 261°
2∠c = 279°
∠c
= 279° ÷ 2
= 139.5°
(b)
From (1)
∠d
= 2∠c - 261°
= 279° - 261°
= 18°
Answer(s): (a) 139.5°; (b) 18°