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

  • A
    $0$
  • B
    $1$
  • C
    $a + b + c$
  • D
    $2(a + b + c)$

Explore More

Similar Questions

If $b = 3, c = 4$ and $B = \frac{\pi}{3}$,then the number of triangles that can be constructed is

In a triangle $ABC$, if $r_1=2 r_2=3 r_3$, then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=$

In a $\triangle ABC$,the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ equals:

The angles $A, B$ and $C$ of a triangle $ABC$ are in $AP$. If $b: c = \sqrt{3}: \sqrt{2}$,then the angle $A$ is (in $^{\circ}$)

The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one,then the sides of the triangle (in units) are

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo