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## In two right triangles ΔABC and ΔDEF,if sinB = sin E then prove that ∠ B = ∠E

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Question.In two right triangles ΔABC and ΔDEF,if sinB = sin E then prove that angle B = angleE

We are given two right triangles ΔABC and ΔDEF in which sinB =sinE

Let ∠A and ∠D are  right triangles of ΔABC and ΔDEF respectively

GIVEN : sinB = sinE

TO PROVE: ∠B =∠E

Proof: In ΔABC and ΔDEF

sinB = sinE

${\color{DarkBlue} \boldsymbol{\frac{BC}{AC}=\frac{EF}{DF}}}$

${\color{DarkBlue} \boldsymbol{\frac{BC}{EF}=\frac{AC}{DF}}}$

${\color{DarkBlue} \boldsymbol{Let},\boldsymbol{\frac{BC}{EF}=\frac{AC}{DF}=k}}$

BC = kEF  & AC = kDF……(i)

Applying Pythagoras theorem in both triangles ΔABC and ΔDEF, we have

${\color{DarkBlue} \boldsymbol{AB=\sqrt{BC^{2} -AC^{2}},DE=\sqrt{EF^{2}-DF^{2}}.....(ii)}}$

Putting the value of BC and AC in the equation  (i)

${\color{DarkBlue} \boldsymbol{AB=\sqrt{k^{2}EF^{2} -k^{2}DF^{2}}}}$

${\color{DarkBlue} \boldsymbol{AB=k\sqrt{EF^{2} -DF^{2}}}}$

Since we have to prove the angles of two triangles, so we are needed to prove both triangles similar to each other.

And proving both triangles similar to each other, here we have to prove all three sides in proportional to each other.

${\color{DarkBlue} \boldsymbol{\frac{AB}{DE}=\frac{k\sqrt{EF^{2}-DF^{2}}}{\sqrt{EF^{2}-DF^{2}}}=k}.....\boldsymbol{(iii)}}$

From (i),(ii) and (iii) we have

${\color{DarkBlue} \boldsymbol{\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}}}$

∴ΔABC ∼ DEF (SSS rule)

Hence ∠B =∠E

OR

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