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(A) Let the angles of $	riangle ABC$ be $A = 45^{\circ}$,$B = 60^{\circ}$,and $C = 180^{\circ} - (45^{\circ} + 60^{\circ}) = 75^{\circ}$.Since the sm

Vedclass · Accepted Answer

$(\sqrt{3}-1) : 1$

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(A) Let the angles of $	riangle ABC$ be $A = 45^{\circ}$,$B = 60^{\circ}$,and $C = 180^{\circ} - (45^{\circ} + 60^{\circ}) = 75^{\circ}$.Since the smallest angle is $A = 45^{\circ}$ and the greatest angle is $C = 75^{\circ}$,the ratio of the smallest side $a$ to the greatest side $c$ is given by the Sine Rule: $\frac{a}{\sin A} = \frac{c}{\sin C}$.Thus,$\frac{a}{c} = \frac{\sin 45^{\circ}}{\sin 75^{\circ}}$.We know $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.Therefore,$\frac{a}{c} = \frac{1/\sqrt{2}}{(\sqrt{3}+1)/(2\sqrt{2})} = \frac{2}{\sqrt{3}+1}$.Rationalizing the denominator: $\frac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{2(\sqrt{3}-1)}{3-1} = \frac{2(\sqrt{3}-1)}{2} = \sqrt{3}-1$.So,the ratio is $(\sqrt{3}-1) : 1$.

If two angles of $\triangle ABC$ are $45^{\circ}$ and $60^{\circ}$,then the ratio of the smallest side to the greatest side is

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