Question
If in a $\triangle\text{ABC},\angle\text{A}=45^\circ,\angle\text{B}=60^{\circ},$ and $\angle\text{C}=75^{\circ},$ find the ratio of its sides.

Answer

$\angle\text{A}=45^{\circ},\angle\text{B}=60^\circ$ and $\angle\text{C}=75^\circ$ Using sing rule, $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k}$ $\frac{\text{a}}{\sin\text{45}}=\frac{\text{b}}{\sin\text{60}}=\frac{\text{c}}{\sin\text{75}}=\text{k}$ $\frac{\text{a}}{\frac{1}{\sqrt{2}}}=\frac{\text{b}}{\frac{\sqrt{3}}{2}}=\frac{\text{c}}{\frac{\sqrt{3}+1}{2\sqrt{2}}}=\text{k}$ $\text{a:b:c}=2:\sqrt{6}:(\sqrt{3}+1)$

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