Find the length of segments FD & FE? Show the work.

Answer:
FD = 8
FE = 21.6
Step-by-step explanation:
By applying Pythagoras theorem in ΔCDF,
CD² = CF² + FD²
(17)² = (15)² + FD²
289 = 225 + FD²
FD = [tex]\sqrt{289-225}[/tex]
= [tex]\sqrt{64}[/tex]
FD = 8 units
Since AB║DE and CD is a transversal line,
∠BCD ≅ ∠CDF [Alternate interior angles]
m∠CDF = m∠BCD = 55°
By using cosine rule in the right triangle CDE,
cos(55)° = [tex]\frac{\text{Adjacent side}}{\text{Hypotenuse}}[/tex] = [tex]\frac{CD}{DE}[/tex]
cos(55)° = [tex]\frac{17}{EF+8}[/tex]
EF + 8 = [tex]\frac{17}{\text{cos}(55)}[/tex]
FE = 29.64 - 8
≈ 21.6 units