In triangleABC, if b=24,c=20 & sin(A/2)=1/?a .find a

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{b}=24 \mathrm{and} \mathrm{c}=20 \\ \sin \left(\frac{\mathrm{A}}{2}\right)=\frac{1}{\sqrt{\mathrm{a}}} \\ \mathrm{Square} \mathrm{both} \mathrm{sides} \\ {\sin }^2\left(\frac{\mathrm{A}}{2}\right)=\frac{1}{\mathrm{a}} \\ \mathrm{Use} \mathrm{cosA}=1-2{\sin }^2\frac{\mathrm{A}}{2} \\ \Rightarrow \frac{1-\mathrm{cosA}}{2}=\frac{1}{\mathrm{a}} \\ \Rightarrow 1-\mathrm{cosA}=\frac{2}{\mathrm{a}} \\ \Rightarrow \mathrm{cosA}=1-\frac{2}{\mathrm{a}} \\ \Rightarrow \frac{{\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2}{2\mathrm{bc}}=\frac{\mathrm{a}-2}{\mathrm{a}} \\ \Rightarrow \frac{576+400-{\mathrm{a}}^2}{2\times 24\times 20}=\frac{\mathrm{a}-2}{\mathrm{a}} \\ \Rightarrow \frac{976-{\mathrm{a}}^2}{960}=\frac{\mathrm{a}-2}{\mathrm{a}} \\ \Rightarrow 976\mathrm{a}-{\mathrm{a}}^3=960\mathrm{a}-1920 \\ \Rightarrow {\mathrm{a}}^3-16\mathrm{a}-1920=0 \\ \mathrm{Solution} \mathrm{of} \mathrm{this} \mathrm{equation} \mathrm{will} \mathrm{give} \mathrm{a}. \end{gathered}$$

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