The figure is not drawn to scale. ∠BAC = 41°, BA = BC and AF//BE. Given that ∠d is
12 of ∠b and ∠d is 4 times of ∠c, find
- ∠d
- ∠a
(a)
d : b = 1 : 2
b = 2 d
c : d = 1 : 4
b = 2 x 4= 8
b : c : d = 8 : 1 : 4
∠BCA = 41° (Isosceles triangle)
∠ACD
= 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°
∠d
= 4 x 1 u
= 4 x 17°
= 68°
(b)
∠BCA = ∠BAC = 41° (Isosceles triangle)
∠CGE = ∠CAF = 68° (Corresponding angles, AF//BE)
∠a
= 68° - 41°
= 27° (Exterior angle of a triangle)
Answer(s): (a) 68°; (b) 27°
