Question

In a $\Delta ABC,$ if a cos A = b cos B, show that the triangle is either isosceles or right - angled.

Answer 1

By the sine rule, we have

$\frac{a}{sinA}=\frac{b}{sinB}=k\left(say\right)$

$\Rightarrow a=ksinAandb=ksinB.$

$\therefore acosA=bcosB$

$\Rightarrow ksinAcosA=ksinBcosB$

$\Rightarrow \frac{1}{2}\left(sin2A\right)=\frac{1}{2}\left(sin2B\right)$

$\Rightarrow sin2A=sin2B$

$\Rightarrow sin2A-sin2B=0$

$\Rightarrow 2cos(A+B)sin(A-B)=0$

$\Rightarrow cos(A+B)=0orsin(A-B)=0$

$\Rightarrow \left(AB\right)=\frac{\pi }{2}orA=B=0$

$\Rightarrow \angle C=\frac{\pi }{2}orA=B.$

Hence, $\Delta ABC,$ is right - angled or isosceles