F.3 maths

1a) AB=BF=2 ( same triangle)
angle ABE+angle EBF= angleFBC+ angle EBF=90˚
so angle ABF=90˚
AF^2=AB^2+BF^2
=4+4
=8
AF=√8

b)tan angle AFB=2/2=1
angle AFB=45˚
AE=FC=3 (same triangle)
AF=√8
EF=1

EF^2+AF^2=1+8=9
AE^2=9

therefore angle EFA=90˚
angle EFB= angle EFA+ angle AFB
=90˚+45˚
=135˚