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 = 47° (Verticallly opposite angles)
∠HEL
= 47° - 12°
= 35°
∠EGL
= 180° - 107°
= 73° (Angles on a straight line)
∠GLK
= 73° + 35°
= 108° (Exterior angle of a triangle)
(b)
∠EFL
= 108° - 47°
= 61° (Exterior angle of a triangle)
Answer(s): (a) 108°; (b) 61°