If two angles of triangle ABC are frac{pi}{4} | vedclass

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(B) Let the angles of $	riangle ABC$ be $A = \frac{\pi}{4} = 45^{\circ}$,$B = \frac{\pi}{3} = 60^{\circ}$,and $C$.The sum of angles in a triangle is

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the angles of $	riangle ABC$ be $A = \frac{\pi}{4} = 45^{\circ}$,$B = \frac{\pi}{3} = 60^{\circ}$,and $C$.The sum of angles in a triangle is $\pi$ radians $(180^{\circ})$.So,$C = 180^{\circ} - (45^{\circ} + 60^{\circ}) = 75^{\circ}$.The angles are $45^{\circ}, 60^{\circ}, 75^{\circ}$.The smallest angle is $45^{\circ}$ and the greatest angle is $75^{\circ}$.By the Sine Rule,$\frac{a}{\sin A} = \frac{c}{\sin C}$,where $a$ is the smallest side and $c$ is the greatest side.The ratio of the smallest side to the greatest side is $\frac{a}{c} = \frac{\sin 45^{\circ}}{\sin 75^{\circ}}$.We know $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.Thus,$\frac{a}{c} = \frac{1/\sqrt{2}}{(\sqrt{3}+1)/(2\sqrt{2})} = \frac{2}{\sqrt{3}+1}$.Rationalizing the denominator: $\frac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{2(\sqrt{3}-1)}{3-1} = \sqrt{3}-1$.Therefore,the ratio is $(\sqrt{3}-1) : 1$.

If two angles of $\triangle ABC$ are $\frac{\pi}{4}$ and $\frac{\pi}{3}$,then the ratio of the smallest and greatest side is

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