The sides $BC,BA$  and $CA$  of $\Delta ABC$  have been produced to $D,E$  and $F$  respectively, as shown in the given figure. Then, the value of $\angle B$   is from the following choices is-

It is given that the figure with some marked angles.
We have to find $\angle B$ .
Consider the figure, vertically opposite angles are equal, so,
$\angle BAC=\angle FAE$ 
$\angle BAC={35}^o$ 
The linear pairs are $\angle BCA,\angle ACD$ .
$\angle BCA+\angle ACD=\angle BCD$ 
Substitute the known values from the figure,
$\angle BCA+110°=180°$ 
$\angle BCA=180°-110°$ 
$=70°$ 
Consider the triangle $ABC$ ,
$\angle A+\angle B+\angle C={180}^{°}$ 
Substitute the known values from the figure,
${35}^{°}+\angle B+{70}^{°}={180}^{°}$ 
${105}^{°}+\angle B={180}^{°}$ 
$\angle B={180}^{°}-{105}^{°}$ 
$\angle B={75}^{°}$ 
The value of $\angle B$  is $75°$ .
Hence option 4 is correct.