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

(A) Let the angles of the triangle be $A = \frac{\pi}{4}$,$B = \frac{\pi}{3}$,and $C = \pi - (\frac{\pi}{4} + \frac{\pi}{3}) = \pi - \frac{7\pi}{12} =

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the angles of the triangle be $A = \frac{\pi}{4}$,$B = \frac{\pi}{3}$,and $C = \pi - (\frac{\pi}{4} + \frac{\pi}{3}) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12}$.Since $\frac{\pi}{4} By the Law of Sines,the ratio of the sides is proportional to the sine of their opposite angles: $\frac{a}{c} = \frac{\sin A}{\sin C}$.$\frac{a}{c} = \frac{\sin(\frac{\pi}{4})}{\sin(\frac{5\pi}{12})} = \frac{\sin(\frac{\pi}{4})}{\sin(\frac{\pi}{4} + \frac{\pi}{6})}$.Using the formula $\sin(x+y) = \sin x \cos y + \cos x \sin y$:$\sin(\frac{5\pi}{12}) = \sin(\frac{\pi}{4})\cos(\frac{\pi}{6}) + \cos(\frac{\pi}{4})\sin(\frac{\pi}{6}) = (\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}})(\frac{1}{2}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$.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 sides is

Similar Questions

The ratio of the sides of triangle $ABC$ is $1:\sqrt{3}:2$. The ratio of angles $A:B:C$ is

In $\triangle ABC$,if $(r_2-r_1)(r_3-r_1)=2 r_2 r_3$,then $2(r+R)=$

In $\triangle ABC$,with usual notations,if $b=3$,$c=8$,and $m\angle A=60^{\circ}$,then the circumradius of the triangle is . . . . . . units.

In triangle $ABC$,$(b + c)\cos A + (c + a)\cos B + (a + b)\cos C = $

In a triangle $ABC$,if $a=5, b=3, c=7$,then $\sqrt{\frac{\sin(A-B)}{\sin(A+B)}}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module