Question

$Ifina\Delta ABC,co{s}^{2}A+co{s}^{2}B+co{s}^{2}C=1,\text{prove that the triangle is right angled.}$

Answer 1

$Let\Delta ABC\text{be any triangle}In\Delta ABCco{s}^{2}A+co{s}^{2}B+co{s}^{2}C=1\Rightarrow co{s}^{2}A+co{s}^{2}B+co{s}^2[\pi -(B+A)]=1(\because A+B+C=\lambda )\Rightarrow co{s}^{2}A+co{s}^{2}B+co{s}^2(B+A)=1\Rightarrow co{s}^{2}A+co{s}^{2}B=1-co{s}^2(B+A)\Rightarrow co{s}^{2}A+co{s}^{2}B=si{n}^2(B+A)\Rightarrow co{s}^{2}A+co{s}^{2}B={\left(sinAcosBcosAsinB\right)}^2\Rightarrow co{s}^{2}A+co{s}^{2}B=si{n}^{2}Aco{s}^{2}B+co{s}^{2}Asi{n}^{2}B+2sinAsinBcosAcosB\Rightarrow 2co{s}^{2}Aco{s}^{2}B=2sinAsinBcosAcosB\Rightarrow cosAcosB=sinAsinB=0cos(A+B)=0cos(A+B)=cos{90}^{\circ }A+B={90}^{\circ }\Rightarrow C={90}^{\circ }(\because A+B+C={180}^{\circ })Hence,\Delta ABC\text{is right angled.}$