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