The figure is not drawn to scale. ∠CBD = 41°, CB = CD and BG//CF. Given that ∠y is
12 of ∠w and ∠y is 4 times of ∠x, find
- ∠y
- ∠v
(a)
y : w = 1 : 2
w = 2 y
x : y = 1 : 4
w = 2 x 4= 8
w : x : y = 8 : 1 : 4
∠CDB = 41° (Isosceles triangle)
∠BDE
= 180° - 41°
= 139° (Angles on a straight line)
1 u + 4 u + 8 u + 139° = 360° (Sum of angles in a quadrilateral)
13 u = 360° - 139° = 221°
1 u = 221 ÷ 13 = 17°
∠y
= 4 x 1 u
= 4 x 17°
= 68°
(b)
∠CDB = ∠CBD = 41° (Isosceles triangle)
∠DHF = ∠DBG = 68° (Corresponding angles, BG//CF)
∠v
= 68° - 41°
= 27° (Exterior angle of a triangle)
Answer(s): (a) 68°; (b) 27°