In a $ riangle A B C$, if $a \cos A=b \cos B$, where $a
eq b$, then $ riangle A B C$ is

Question

In a $	riangle A B C$, if $a \cos A=b \cos B$, where $a 
eq b$, then $	riangle A B C$ is

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { } a \cos A=b \cos B \ & a\left(\frac{b^2+c^2-a^2}{2 b c}ight)=b\left(\frac{a^2+c^2-b^2}{2 a c}ight) \ & \Rightarrow \frac{a}{b}\left(b^2+c^2-a^2ight)=\frac{b}{a}\left(a^2+c^2-b^2ight) \ & \Rightarrow a^2\left(b^2+c^2-a^2ight)=b^2\left(a^2+c^2-b^2ight)\end{aligned}$ $ \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ight)\left(a^2+b^2ight)+c^2\left(b^2-a^2ight)=0 \ & \Rightarrow \quad\left(a^2-b^2ight)\left[a^2+b^2-c^2ight]=0 \ & 	ext { As }(a 
eq b), a^2-b^2 
eq 0 \ & \quad a^2+b^2-c^2=0 \Rightarrow a^2+b^2=c^2 \end{aligned} $ $	herefore 	riangle A B C$ is a right angled triangle with $A B$ as hypotenuse

In a $\triangle A B C$, if $a \cos A=b \cos B$, where $a \neq b$, then $\triangle A B C$ is

Solution