frac{acos A + bcos B + ccos C}{a + b + c} = | vedclass

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

Question

(B) We know that $a = 2R\sin A$,$b = 2R\sin B$,and $c = 2R\sin C$.Substituting these in the expression:$\frac{a\cos A + b\cos B + c\cos C}{a + b + c}

Vedclass · Accepted Answer

$r/R$

Vedclass · Answer

(B) We know that $a = 2R\sin A$,$b = 2R\sin B$,and $c = 2R\sin C$.Substituting these in the expression:$\frac{a\cos A + b\cos B + c\cos C}{a + b + c} = \frac{2R\sin A \cos A + 2R\sin B \cos B + 2R\sin C \cos C}{2R\sin A + 2R\sin B + 2R\sin C}$$= \frac{\sin 2A + \sin 2B + \sin 2C}{2(\sin A + \sin B + \sin C)}$Using the identity $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$ and $\sin A + \sin B + \sin C = 4\cos(A/2)\cos(B/2)\cos(C/2)$:$= \frac{4\sin A \sin B \sin C}{2(4\cos(A/2)\cos(B/2)\cos(C/2))}$$= \frac{4(2\sin(A/2)\cos(A/2))(2\sin(B/2)\cos(B/2))(2\sin(C/2)\cos(C/2))}{8\cos(A/2)\cos(B/2)\cos(C/2)}$$= 8\sin(A/2)\sin(B/2)\sin(C/2)$Since $r = 4R\sin(A/2)\sin(B/2)\sin(C/2)$,we have $4\sin(A/2)\sin(B/2)\sin(C/2) = r/R$.Thus,the expression equals $r/R$.

$\frac{a\cos A + b\cos B + c\cos C}{a + b + c} = $

Similar Questions

If $H$ is the orthocentre of $\triangle ABC$ and $AH=x, BH=y, CH=z$,then $\frac{abc}{xyz}=$

In a $\triangle ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then $\triangle ABC$ is

If $b+c=3a$,then $\cot \frac{B}{2} \cot \frac{C}{2}$ is equal to :

If the median of $\Delta ABC$ through $A$ is perpendicular to $AB$,then

The inradius of the triangle whose sides are $3, 5, 6$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module