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 = 39° (Verticallly opposite angles)
∠HEL
= 39° - 12°
= 27°
∠EGL
= 180° - 99°
= 81° (Angles on a straight line)
∠GLK
= 81° + 27°
= 108° (Exterior angle of a triangle)
(b)
∠EFL
= 108° - 39°
= 69° (Exterior angle of a triangle)
Answer(s): (a) 108°; (b) 69°
