The figure is not drawn to scale. ∠BAC = 41°, BA = BC and AF//BE. Given that ∠z is
13 of ∠x and ∠z is 4 times of ∠y, find
- ∠z
- ∠w
(a)
z : x = 1 : 3
x = 3 z
y : z = 1 : 4
x = 3 x 4= 12
x : y : z = 12 : 1 : 4
∠BCA = 41° (Isosceles triangle)
∠ACD
= 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°
∠z
= 4 x 1 u
= 4 x 13°
= 52°
(b)
∠BCA = ∠BAC = 41° (Isosceles triangle)
∠CGE = ∠CAF = 52° (Corresponding angles, AF//BE)
∠w
= 52° - 41°
= 11° (Exterior angle of a triangle)
Answer(s): (a) 52°; (b) 11°
