Question

Pls help nothings helping !!

Answer 1

Answer:

Step-by-step explanation:

∠BCD=∠DEF=38°(alternate angles)

∠FDE=90-38=52°

Δ BCD≅ΔFED

$\frac{CD}{sin~90} =\frac{10}{sin~38} \\CD=\frac{10}{sin~38} \\\frac{DE}{sin ~90} =\frac{17}{sin ~52} \\DE=\frac{17}{sin~52} \\CE=CD+DE=\frac{10}{sin~38} +\frac{17}{sin~52} \\=?$