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# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Let $a,b,c$ be the sides of a triangle and $A,B,C$ be the angles opposite to those sides respectively. If $\sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B$, then prove that the triangle is isosceles.

I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 6 from 2016

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Do you really need a hint? Try it first!

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Let $a,b,c$ be the sides of a triangle and $A,B,C$ be the angles opposite to those sides respectively. Given $\sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B$ $\Rightarrow \sin A\cos B- \cos A \sin B=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B$ $\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B$

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$\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B$ $\Rightarrow \sin A\cos B(1-\frac{a}{a+b})=\cos A \sin B(1-\frac{b}{a+b})$ $\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B$

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$\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B$ $\Rightarrow b\sin A\cos B=a\cos A \sin B$ $\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A}$

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$\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A}$ $\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{b}{\sin B}$  [since $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$] $\Rightarrow \frac{\cos B}{\cos A}=1$ $\cos B=\cos A$ $A=B$.

$\Rightarrow \Delta ABC$ is isosceles  (Proved).

# Connected Program at Cheenta

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.