Question

Answer each of the following questions in one word or one sentence or as per exact requirement of for question:

$\text{If the sides of a triangle are proportional to}2,\sqrt{6}and\sqrt{3}-1,\text{find the measure of its greatest angle.}$

Answer 1

$Leta=2,b=\sqrt{6},c=\sqrt{3}-1cosA=\frac{b^2+c^2-a^2}{2bc}\Rightarrow cosA=\frac{6+(3+1-2\sqrt{3})-4}{2\times \sqrt{6}\times (\sqrt{3}-1)}\Rightarrow cosA=\frac{6-2\sqrt{3}}{2\times \sqrt{6}\times (\sqrt{3}-1)}\Rightarrow cosA=\frac{2\sqrt{3}(\sqrt{3}-1)}{2\sqrt{6}(\sqrt{3}-1)}\Rightarrow cosA=\frac{1}{\sqrt{2}}A=45.cosC=\frac{a^2+b^2-c^2}{2ab}\Rightarrow cosA=\frac{4+6-(3+1-2\sqrt{3})}{2\times 2\times \sqrt{6}}\Rightarrow cosA=\frac{6+2\sqrt{3}}{2\times 2\times \sqrt{6}}\Rightarrow cosA=\frac{2\sqrt{3}(\sqrt{3}+1)}{4\sqrt{6}}=cosA=\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)A=15\angle C=180-(A+B)=180-(45+15)={120}^{\circ }$