Answer:
dude your teacher is on crack
Substitute the variable terms with the coordinates and the equation is
3=1/3(5)+b
solve for b
You can calculate it using the law of cosines: c^2=a^2+b^2-2*a*b*cos(C)
your triangle is
CD=15=a
CE=?=b
DE=CE+3=b+3=c
and C=90°
-> insert those values, with c substituted with b+3 to remove c
c^2=a^2+b^2-2*a*b*cos(C)
(b+3)^2=15^2+b^2-2*15*b*cos(90)
cos(90)=0->
(b+3)^2=15^2+b^2
b^2+2*3*b+3^2=225+b^2
6b+9=225
6b=216
b=36=CE
DE=CE+3=36+3=39
Answer: 21
Step-by-step explanation:
