# In triangles ABC and DEF, ∠A = ∠E = 40°, AB : ED = AC : EF and ∠F = 65°, then ∠B =

Question:

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) $85^{\circ}$

Solution:

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 $(c)$.