Mathematics - Properties of Triangles Question with Solution | TestHub

Let A, B and $C$ be the three angles of $\Delta ABC$ and Let $a=10$ and $b=9$ It is given that the angles are in $\mathrm{AP}$. $\therefore 2\mathrm{B}=\mathrm{A}+\mathrm{C}$ on adding $\mathrm{B}$ both the sides, we get $$\begin{array}{l}3\mathrm{B}=\mathrm{A}+\mathrm{B}+\mathrm{C}\\ \Rightarrow 3\mathrm{B}=180^{\circ }\Rightarrow \mathrm{B}=60^{\circ }\end{array}$$ Now, we know $\cos \mathrm{B}=\frac{{\mathrm{a}}^2+{\mathrm{c}}^2-{\mathrm{b}}^2}{2\mathrm{ac}}$ $$\begin{array}{l}\Rightarrow \cos 60^{\circ }=\frac{10^2+{\mathrm{c}}^2-9^2}{2\times 10\times \mathrm{c}}\\ \Rightarrow \frac{1}{2}=\frac{100+{\mathrm{c}}^2-81}{20\mathrm{c}}\\ \Rightarrow {\mathrm{c}}^2-10\mathrm{c}+19=0\Rightarrow \mathrm{c}=5\pm \sqrt{6}\end{array}$$