The figure is not drawn to scale. ∠DCE = 41°, DC = DE and CH//DG. Given that ∠k is
13 of ∠i and ∠k is 4 times of ∠j, find
- ∠k
- ∠h
(a)
k : i = 1 : 3
i = 3 k
j : k = 1 : 4
i = 3 x 4= 12
i : j : k = 12 : 1 : 4
∠DEC = 41° (Isosceles triangle)
∠CEF
= 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°
∠k
= 4 x 1 u
= 4 x 13°
= 52°
(b)
∠DEC = ∠DCE = 41° (Isosceles triangle)
∠EJG = ∠ECH = 52° (Corresponding angles, CH//DG)
∠h
= 52° - 41°
= 11° (Exterior angle of a triangle)
Answer(s): (a) 52°; (b) 11°
