With usual notations,the perimeter of a trian | vedclass

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

Question

(D) Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Then $\sin A = \frac{a}{k}$,$\sin B = \frac{b}{k}$,and $\sin C = \frac{c}{k}$. T

Vedclass · Accepted Answer

$\frac{\pi^c}{6}$

Vedclass · Answer

(D) Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Then $\sin A = \frac{a}{k}$,$\sin B = \frac{b}{k}$,and $\sin C = \frac{c}{k}$. The perimeter of the triangle is $a+b+c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$. According to the problem,$a+b+c = 6 	imes \left( \frac{\sin A + \sin B + \sin C}{3} ight)$. Substituting the values,we get $a+b+c = 2(\sin A + \sin B + \sin C) = 2 \left( \frac{a+b+c}{k} ight)$. Thus,$1 = \frac{2}{k}$,which implies $k = 2$. Since $\sin A = \frac{a}{k}$ and $a=1$,we have $\sin A = \frac{1}{2}$. Therefore,$A = \frac{\pi^c}{6}$.

With usual notations,the perimeter of a triangle $ABC$ is $6$ times the arithmetic mean of the sines of its angles. If $a=1$,then the measure of angle $A$ is:

Similar Questions

The sides of a triangle are $\sin \alpha$,$\cos \alpha$,and $\sqrt{1 + \sin \alpha \cos \alpha}$ for some $0 < \alpha < \frac{\pi}{2}$. Then the greatest angle of the triangle is.....$^o$

In a triangle $ABC$,if $B = 3C$,then the values of $\sqrt{\frac{b + c}{4c}}$ and $\frac{b - c}{2c}$ are

If in a $\triangle ABC$, $r_1=2$, $r_2=3$ and $r_3=6$, then $a$ equals to

The angles of a triangle are in the ratio $1:3:5$. Find the greatest angle. (in $\pi /9$)

With usual notations in $\Delta ABC$,$a=3$,$c=2$ and $\sin C=\frac{2}{3}$,then $\angle A=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module