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
- ∠t
- ∠v
(a)
∠CKE = 180° - ∠t (Interior angles, KC//HD)
∠FEK = 180° - ∠t (Angles on a straight line)
∠FKE = 180° - ∠t (Isosceles triangle)
58° + 180° - ∠t + 180° - ∠t + ∠v + 15° = 180° (Angles on a straight line, BH)
58° + 180° + 180° + 15° - ∠t - ∠t + ∠v = 180°
433° - 2∠t + ∠v = 180°
2∠t - ∠v = 433° - 180°
2∠t - ∠v = 253°
∠v = 2∠t - 253° --- (1)
∠KFG
= ∠KGF
= 2 x (180° - ∠t)
= 360° - 2∠t (Exterior angle of a triangle)
∠v = 180° - (360° - 2∠t) - (360° - 2∠t)
∠v = 180° - 360° + 2∠t - 360° + 2∠t
∠v = 180° - 360° - 360° + 2∠t + 2∠t
∠v = 4∠t - 540° (Angles sum of triangle)
∠v = 4∠t - 540° --- (2)
(2) = (1)
4∠t - 540° = 2∠t - 253°
4∠t - 2∠t= 540° - 253°
2∠t = 287°
∠t
= 287° ÷ 2
= 143.5°
(b)
From (1)
∠v
= 2∠t - 253°
= 287° - 253°
= 34°
Answer(s): (a) 143.5°; (b) 34°