Mathematics - Properties of Triangle Question with Solution | TestHub

If the angles $A,B,C$ are in arithmetic progression, then $A+C=2B$. Since $A+B+C=\pi$, we have $2B+B=\pi$, so $3B=\pi$, which means $B=\pi /3=60^{\circ }$. A. The cosine rule for angle $B$ is $b^2=a^2+c^2-2ac\cos B$. Since $B=60^{\circ }$, $\cos B=\cos 60^{\circ }=1/2$. Thus $b^2=a^2+c^2-2ac(1/2)=a^2+c^2-ac$. So A is correct. B. The formula $r=(s-b)\tan (B/2)$ is a standard identity for the inradius of any triangle. So B is correct. C. By the sine rule, $b/\sin B=2R$. Since $B=60^{\circ }$, $b/(\sin 60^{\circ })=2R⟹b/(\sqrt{3}/2)=2R⟹2b/\sqrt{3}=2R⟹R=b/\sqrt{3}$. So C is correct. D. The condition $a+c=2b$ implies that the sides are in arithmetic progression. Angles in AP does not necessarily imply sides in AP. For example, a $30^{\circ }-60^{\circ }-90^{\circ }$ triangle has angles in AP, but its sides are in ratio $1:\sqrt{3}:2$, which are not in AP. So D is incorrect.