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