The figure is not drawn to scale. ∠DCE = 41°, DC = DE and CH//DG. Given that ∠v is
12 of ∠s and ∠v is 4 times of ∠t, find
- ∠v
- ∠r
(a)
v : s = 1 : 2
s = 2 v
t : v = 1 : 4
s = 2 x 4= 8
s : t : v = 8 : 1 : 4
∠DEC = 41° (Isosceles triangle)
∠CEF
= 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°
∠v
= 4 x 1 u
= 4 x 17°
= 68°
(b)
∠DEC = ∠DCE = 41° (Isosceles triangle)
∠EJG = ∠ECH = 68° (Corresponding angles, CH//DG)
∠r
= 68° - 41°
= 27° (Exterior angle of a triangle)
Answer(s): (a) 68°; (b) 27°
