The figure is not drawn to scale. ∠CBD = 41°, CB = CD and BG//CF. Given that ∠d is
13 of ∠b and ∠d is 4 times of ∠c, find
- ∠d
- ∠a
(a)
d : b = 1 : 3
b = 3 d
c : d = 1 : 4
b = 3 x 4= 12
b : c : d = 12 : 1 : 4
∠CDB = 41° (Isosceles triangle)
∠BDE
= 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°
∠d
= 4 x 1 u
= 4 x 13°
= 52°
(b)
∠CDB = ∠CBD = 41° (Isosceles triangle)
∠DHF = ∠DBG = 52° (Corresponding angles, BG//CF)
∠a
= 52° - 41°
= 11° (Exterior angle of a triangle)
Answer(s): (a) 52°; (b) 11°
