Mathematics - Properties of Triangle Question with Solution | TestHub

The condition $a\cos A+b\cos B=c\cos C$ is equivalent to $2R\sin A\cos A+2R\sin B\cos B=2R\sin C\cos C$, which simplifies to $\sin 2A+\sin 2B=\sin 2C$. Using $A+B=\pi -C$, this becomes $2\sin (A+B)\cos (A-B)=2\sin C\cos C$, which implies $\cos (A-B)=\cos C$. This means $A-B=C$ or $B-A=C$, both leading to the triangle being right-angled (at $A$ or $B$). Assume $A=90^{\circ }$. Then $a$ is the hypotenuse, $R=a/2$, $r=(b+c-a)/2$, $r_a=(a+b+c)/2$, $r_b=(a+c-b)/2$, $r_c=(a+b-c)/2$. A) $r_b+r_c=(a+c-b)/2+(a+b-c)/2=2a/2=a$. (Correct) B) $r+r_a=(b+c-a)/2+(a+b+c)/2=(2b+2c)/2=b+c$. (Correct) C) $R=a/2$. $r_a-(r_b+r_c)=(a+b+c)/2-a=(-a+b+c)/2$. Thus $R=r_a-(r_b+r_c)$ implies $a=-a+b+c$, or $2a=b+c$, not generally true. (Incorrect) D) The identity $r_a+r_b+r_c-r=4R$ means $r_a+r_b+r_c-4R=r$. So option D states $s=r$. For $A=90^{\circ }$, $s=(a+b+c)/2$ and $r=(b+c-a)/2$. $s=r$ implies $a=0$, which is impossible. (Incorrect)