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 = 42° (Verticallly opposite angles)
∠HEL
= 42° - 15°
= 27°
∠EGL
= 180° - 95°
= 85° (Angles on a straight line)
∠GLK
= 85° + 27°
= 112° (Exterior angle of a triangle)
(b)
∠EFL
= 112° - 42°
= 70° (Exterior angle of a triangle)
Answer(s): (a) 112°; (b) 70°