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