The figure is not drawn to scale. ∠XWY = 41°, XW = XY and WB//XA. Given that ∠b is
13 of ∠z and ∠b is 4 times of ∠a, find
- ∠b
- ∠y
(a)
b : z = 1 : 3
z = 3 b
a : b = 1 : 4
z = 3 x 4= 12
z : a : b = 12 : 1 : 4
∠XYW = 41° (Isosceles triangle)
∠WYZ
= 180° - 41°
= 139° (Angles on a straight line)
1 u + 4 u + 12 u + 139° = 360° (Sum of angles in a quadrilateral)
17 u = 360° - 139° = 221°
1 u = 221 ÷ 17 = 13°
∠b
= 4 x 1 u
= 4 x 13°
= 52°
(b)
∠XYW = ∠XWY = 41° (Isosceles triangle)
∠YCA = ∠YWB = 52° (Corresponding angles, WB//XA)
∠y
= 52° - 41°
= 11° (Exterior angle of a triangle)
Answer(s): (a) 52°; (b) 11°