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

(A) The perimeter of the triangle is $P = a + b + c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.Given $a

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) The perimeter of the triangle is $P = a + b + c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.Given $a + b + c = 6 	imes \frac{\sin A + \sin B + \sin C}{3} = 2(\sin A + \sin B + \sin C)$.Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,where $R$ is the circumradius.Substituting these into the equation: $2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.This implies $2R = 2$,so $R = 1$.Since $a = 2R \sin A$ and $a = 1$,we have $1 = 2(1) \sin A$.Therefore,$\sin A = \frac{1}{2}$,which gives $A = \frac{\pi}{6}$ or $A = \frac{5\pi}{6}$.Given the standard context of such problems,we select $A = \frac{\pi}{6}$.

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 $\angle A=$

Similar Questions

Let $ABC$ be a triangle such that $AB=15$ and $AC=9$. The bisector of $\angle BAC$ meets $BC$ in $D$. If $\angle ACB=2\angle ABC$,then $BD$ is

If the angles of a triangle $ABC$ are in the ratio $1: 2: 3$,then the corresponding sides are in the ratio

If in a $\triangle ABC$, $r_1 < r_2 < r_3$, then:

In $\Delta ABC$,$({b^2} - {c^2})\cot A + ({c^2} - {a^2})\cot B + ({a^2} - {b^2})\cot C = $

In a $\triangle ABC$,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module