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In a $\triangle A B C$, if $a \cos A=b \cos B$, where $a \neq b$, then $\triangle A B C$ is
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Verified Answer
The correct answer is:
Right angled triangle
$\begin{aligned} & \text { } a \cos A=b \cos B \\ & a\left(\frac{b^2+c^2-a^2}{2 b c}\right)=b\left(\frac{a^2+c^2-b^2}{2 a c}\right) \\ & \Rightarrow \frac{a}{b}\left(b^2+c^2-a^2\right)=\frac{b}{a}\left(a^2+c^2-b^2\right) \\ & \Rightarrow a^2\left(b^2+c^2-a^2\right)=b^2\left(a^2+c^2-b^2\right)\end{aligned}$
$$
\begin{aligned}
& \Rightarrow \quad a^2 b^2+a^2 c^2-a^4=a^2 b^2+b^2 c^2-b^4 \\
& \Rightarrow \quad a^4-b^4+b^2 c^2-a^2 c^2=0 \\
& \Rightarrow \quad\left(a^2-b^2\right)\left(a^2+b^2\right)+c^2\left(b^2-a^2\right)=0 \\
& \Rightarrow \quad\left(a^2-b^2\right)\left[a^2+b^2-c^2\right]=0 \\
& \text { As }(a \neq b), a^2-b^2 \neq 0 \\
& \quad a^2+b^2-c^2=0 \Rightarrow a^2+b^2=c^2
\end{aligned}
$$
$\therefore \triangle A B C$ is a right angled triangle with $A B$ as hypotenuse.
$$
\begin{aligned}
& \Rightarrow \quad a^2 b^2+a^2 c^2-a^4=a^2 b^2+b^2 c^2-b^4 \\
& \Rightarrow \quad a^4-b^4+b^2 c^2-a^2 c^2=0 \\
& \Rightarrow \quad\left(a^2-b^2\right)\left(a^2+b^2\right)+c^2\left(b^2-a^2\right)=0 \\
& \Rightarrow \quad\left(a^2-b^2\right)\left[a^2+b^2-c^2\right]=0 \\
& \text { As }(a \neq b), a^2-b^2 \neq 0 \\
& \quad a^2+b^2-c^2=0 \Rightarrow a^2+b^2=c^2
\end{aligned}
$$
$\therefore \triangle A B C$ is a right angled triangle with $A B$ as hypotenuse.
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