Question

Let D be any point on the base of isosceles triangle ABC. AC is extended to E so that CE=CD. ED is extended to meet AB at F. Angle CED = 10°.... Find cosine of angle AFD.

Answer 1

(click on the above image for diagram)

CD = CE

angle CDE = angle CED = 10°

By applying angle sum property of ∆ECD

angle ECD = 160°

angle ACB = 180° - angle ECD

angle ACB = 180° - 160° = 20°

AB = AC (∆ABC is Isosceles)

angle ACB = angle ABC = 20°

By applying angle sum property of ∆ABC

angle BAC = 140°

By applying angle sum property of ∆AFE

angle AFE = 30°

i.e.

angle AFD = 30°

cos(angle AFD) = cos30°

cos(angle AFD) = √3/2

Hence, cosine of angle AFD is √3/2