Question

$Ina\Delta ABC,ifcosC=\frac{sinA}{2sinB},\text{prove that the triangle is isosceles.}$

Answer 1

$Let\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=k.Then,sinA=ka,sinB=kb,sinC=kcNow,cosC=\frac{sinA}{2sinB}2sinBcosC=sinA2\left(\frac{a^2+b^2-c^2}{2ab}\right)kb=ka{a}^2+b^2-c^2=a^2{b}^2=c^{2}b=c\Delta ABC\text{is isosceles.}P.Q.\frac{cosA}{bcosC+ccosB}+\frac{cosB}{ccosA+acosC}+\frac{cosC}{acosB+bcosA}=\frac{a^2+b^2+c^2}{2abc}Inany\Delta ABC,\text{we have}a=bcosC+ccosBb=ccosA+acosCc=acosB+bcosA\text{Therefore,}LHS=\frac{cosA}{bcosC+ccosB}+\frac{cosB}{ccosA+acosC}+\frac{cosC}{acosB+bcosA}=\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}=\frac{b^2+c^2-a^2}{2abc}+\frac{a^2+c^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}=\frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{2abc}=\frac{a^2+b^2-c^2}{2abc}=RHS\text{Hence proved.}PQ.Ina\Delta ABC,ifcosA=sinB-cosC,\text{show that the triangle is a right angled triangle.}cosA=sinB-cosC\Rightarrow cosA+cosC=sinB\Rightarrow 2cos\left(\frac{A+C}{2}\right)cos\left(\frac{A-C}{2}\right)=2sin\frac{B}{2}.cos\frac{B}{2}\Rightarrow cos(\frac{\pi }{2}-\frac{B}{2})cos\left(\frac{A-C}{2}\right)-sin\frac{B}{2}.cos\frac{B}{2}\Rightarrow sin\frac{B}{2}cos\left(\frac{A-C}{2}\right)=sin\frac{B}{2}.cos\frac{B}{2}\Rightarrow cos\left(\frac{A-C}{2}\right)=cos\frac{B}{2}\left[\text{Dividing both sides by sin}\frac{B}{2}\right]\Rightarrow \left(\frac{A-C}{2}\right)=\frac{B}{2}\left[\text{Removing sine from both sides}\right]\Rightarrow A-C=B\Rightarrow A=B+C...\left(i\right)Also,A+B+C=\pi ...\left(ii\right)\text{Solving (i) and (ii), we get}\angle A=\frac{\pi }{2}\therefore \Delta ABC\text{is a right angled triangle.}$