If the sides of a triangle are in the ratio 2: root 6: (root 3)+1 , then the largest angle of the triangle will be? Ans is 75 deg

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$$\begin{gathered} \mathrm{As} \mathrm{sides} \mathrm{are} \mathrm{in} \mathrm{the} \mathrm{ratio} 2:\sqrt{6}:\sqrt{3}+1 \\ \mathrm{Let} \mathrm{a}=2 \mathrm{units} ,\mathrm{b}=\sqrt{6} \mathrm{units} \mathrm{and} \mathrm{c}=\left(\sqrt{3}+1\right) \mathrm{units} \\ 2<\sqrt{6}<\sqrt{3}+1 \\ \mathrm{Hence} \mathrm{c} \mathrm{is} \mathrm{the} \mathrm{greatest} \mathrm{side}. \\ \mathrm{Now} \mathrm{greatest} \mathrm{angle} \mathrm{is} \mathrm{opposite} \mathrm{to} \mathrm{greatest} \mathrm{side}. \\ \mathrm{By} \mathrm{Cosine} \mathrm{rule}, \mathrm{we} \mathrm{have}, \\ \mathrm{cosC}=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}}{2\mathrm{ab}} \\ =\frac{2^2+{\left(\sqrt{6}\right)}^2-{\left(\sqrt{3}+1\right)}^2}{2.2.\sqrt{6}} \\ =\frac{4+6-\left(3+1+2\sqrt{3}\right)}{4\sqrt{6}} \\ =\frac{10-4-2\sqrt{3}}{4\sqrt{6}} \\ =\frac{6-2\sqrt{3}}{4\sqrt{6}} \\ \mathrm{Dividing} \mathrm{each} \mathrm{term} \mathrm{by} 2\sqrt{3} \mathrm{we} \mathrm{get}, \\ \cos \mathrm{C}=\frac{\sqrt{3}-1}{2\sqrt{2}} \\ =\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}} \\ =\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{2}}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right) \\ =\cos 30°.\cos 45°-\sin 30°.\sin 45° \\ \mathrm{Using} \mathrm{identity} \cos \left(\mathrm{A}+\mathrm{B}\right)=\cos \mathrm{A}.\cos \mathrm{B}-\sin \mathrm{A}.\sin \mathrm{B}, \mathrm{we} \mathrm{get}, \\ \cos \mathrm{C}=\cos \left(30°+45°\right)=\cos 75° \\ \Rightarrow \mathrm{C}=75° \\ \mathbf{Hence}\mathbf{ }\mathbf{greatest}\mathbf{ }\mathbf{angle}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{75}\mathbf{°}\mathbf{.}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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