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
- ∠v
- ∠w
(a)
∠CKE = 180° - ∠v (Interior angles, KC//HD)
∠FEK = 180° - ∠v (Angles on a straight line)
∠FKE = 180° - ∠v (Isosceles triangle)
49° + 180° - ∠v + 180° - ∠v + ∠w + 21° = 180° (Angles on a straight line, BH)
49° + 180° + 180° + 21° - ∠v - ∠v + ∠w = 180°
430° - 2∠v + ∠w = 180°
2∠v - ∠w = 430° - 180°
2∠v - ∠w = 250°
∠w = 2∠v - 250° --- (1)
∠KFG
= ∠KGF
= 2 x (180° - ∠v)
= 360° - 2∠v (Exterior angle of a triangle)
∠w = 180° - (360° - 2∠v) - (360° - 2∠v)
∠w = 180° - 360° + 2∠v - 360° + 2∠v
∠w = 180° - 360° - 360° + 2∠v + 2∠v
∠w = 4∠v - 540° (Angles sum of triangle)
∠w = 4∠v - 540° --- (2)
(2) = (1)
4∠v - 540° = 2∠v - 250°
4∠v - 2∠v= 540° - 250°
2∠v = 290°
∠v
= 290° ÷ 2
= 145°
(b)
From (1)
∠w
= 2∠v - 250°
= 290° - 250°
= 40°
Answer(s): (a) 145°; (b) 40°