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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

If in $\triangle ABC$,with usual notations,$a \cdot \cos^2 \frac{C}{2} + c \cdot \cos^2 \frac{A}{2} = \frac{3b}{2}$,then

In $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then

In a $\triangle ABC$,$A = 30^{\circ} + C$ and $R = (\sqrt{3} + 1)r$,where $r$ is the inradius and $R$ is the circumradius. Then:

In a triangle $ABC$,if $r_1 = 2r_2 = 3r_3$,then $\sin A : \sin B : \sin C =$

If in a $\triangle ABC$,$r_3 = r_1 + r_2 + r$,then $\angle A + \angle B$ is equal to (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module