Answer:
Step-by-step explanation:
∠BCD=∠DEF=38°(alternate angles)
∠FDE=90-38=52°
Δ BCD≅ΔFED
[tex]\frac{CD}{sin~90} =\frac{10}{sin~38} \\CD=\frac{10}{sin~38} \\\frac{DE}{sin ~90} =\frac{17}{sin ~52} \\DE=\frac{17}{sin~52} \\CE=CD+DE=\frac{10}{sin~38} +\frac{17}{sin~52} \\=?[/tex]
