Question

In $△ABC,$ if $acosA=bcosB$, then prove that the triangle is either a right-angled or an isosceles triangle.

Answer 1

$acosA=b\sin B-\left(1\right)$

$sinerule=>\frac{a}{\sin A}=\frac{b}{\sin B}=k\left(let\right)$

$=>a=k\sin A;b=k\sin B$

$=>puttingin\left(1\right)$

$=>k\sin AcosA=k\sin BcosB$

$=>\sin AcosA=\sin BcosB$

$(\sin 2\theta =2\sin \theta cos\theta )$

$=>\frac{\sin 2A}{2}=\frac{\sin 2B}{2}$

$=>\sin 2A-\sin 2B=0$

$\sin 2A-\sin 2B=2cos\left(\frac{2A+2B}{2}\right)\sin \left(\frac{2A-2B}{2}\right)=0$

$=>2cos(A+B)\sin (A-B)=0-\left(2\right)$

$=>Eithercos(A+B)=0$

$=>A+B={90}^{\circ }$

$\therefore C=180-A-B$

$={90}^{\circ }$

$\therefore △ABCwillberightangledatC$

$or$

$\sin (A-B)=0$

$=>A-B=0$

$=>A=B$

$\therefore △ABCisisoscelesas2anglesareequal.$

$Henceproved.$