In triangle ABC,if A: B: C = 5: 1: 6,then a: | vedclass

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(A) Given the ratio of angles $A: B: C = 5: 1: 6$. Let the angles be $5k, k, 6k$. Since the sum of angles in a triangle is $180^{\circ}$,we have $5k +

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$\sqrt{3}+1: \sqrt{3}-1: 2\sqrt{2}$

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(A) Given the ratio of angles $A: B: C = 5: 1: 6$. Let the angles be $5k, k, 6k$. Since the sum of angles in a triangle is $180^{\circ}$,we have $5k + k + 6k = 180^{\circ}$,which gives $12k = 180^{\circ}$,so $k = 15^{\circ}$.Thus,$A = 75^{\circ}$,$B = 15^{\circ}$,and $C = 90^{\circ}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.Therefore,$a: b: c = \sin 75^{\circ}: \sin 15^{\circ}: \sin 90^{\circ}$.We know $\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4}$ and $\sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) = \frac{\sqrt{6} - \sqrt{2}}{4}$.Also,$\sin 90^{\circ} = 1$.So,$a: b: c = \frac{\sqrt{6} + \sqrt{2}}{4}: \frac{\sqrt{6} - \sqrt{2}}{4}: 1$.Multiplying by $4$,we get $a: b: c = (\sqrt{6} + \sqrt{2}): (\sqrt{6} - \sqrt{2}): 4$.Dividing by $\sqrt{2}$,we get $a: b: c = (\sqrt{3} + 1): (\sqrt{3} - 1): 2\sqrt{2}$.

In $\triangle ABC$,if $A: B: C = 5: 1: 6$,then $a: b: c =$

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