In triangles $\mathrm{ABC}$ and $\mathrm{DEF}, \angle \mathrm{A}=\angle \mathrm{E}=40^{\circ}, \mathrm{AB}: \mathrm{ED}=\mathrm{AC}: \mathrm{EF}$ and $\angle \mathrm{F}=65^{\circ}$, then $\angle \mathrm{B}=$
(a) $35^{\circ}$
(b) $65^{\circ}$
(c) $75^{\circ}$
(d) $85^{\circ}$
Given: In ΔABC and ΔDEF
$\angle \mathrm{A}=\angle \mathrm{E}=40^{\circ}$
$\mathrm{AB}: \mathrm{ED}=\mathrm{AC}: \mathrm{EF}$
$\angle \mathrm{F}=65^{\circ}$
To find: Measure of angle B.
In ΔABC and ΔDEF
$\angle \mathrm{A}=\angle \mathrm{E}=40^{\circ}$
$\mathrm{AB}: \mathrm{ED}=\mathrm{AC}: \mathrm{EF}$
$\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}(\mathrm{S} . \mathrm{A} . \mathrm{S}$ Similarity crieteria)
Hence in similar triangles ΔABC and ΔDEF
$\angle \mathrm{A}=\angle \mathrm{E}=40^{\circ}$
$\angle \mathrm{C}=\angle \mathrm{F}=65^{\circ}$
$\angle \mathrm{B}=\angle \mathrm{D}$
We know that sum of all the angles of a triangle is equal to $180^{\circ}$.
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
$40^{\circ}+\angle B+65^{\circ}=180^{\circ}$
$\angle \mathrm{B}+115^{\circ}=180^{\circ}$
$\angle \mathrm{B}=180^{\circ}-115^{\circ}$
$\angle \mathrm{B}=75^{\circ}$
Hence the correct answer is $(c)$.