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