In the figure, ABC and CDE are isosceles triangles where AC = BC and CD = CE. Given that DF, EH and AK are straight lines, find
- ∠GLK
- ∠EFL
(a)
∠ABC = ∠BAC = ∠CDE = ∠CED (Isosceles triangle)
∠CED = ∠FEL = 36° (Verticallly opposite angles)
∠HEL
= 36° - 14°
= 22°
∠EGL
= 180° - 106°
= 74° (Angles on a straight line)
∠GLK
= 74° + 22°
= 96° (Exterior angle of a triangle)
(b)
∠EFL
= 96° - 36°
= 60° (Exterior angle of a triangle)
Answer(s): (a) 96°; (b) 60°