If $a=2, b=3, c=4$ in a triangle $ABC$,then $\cos C=$

  • A
    $\frac{1}{4}$
  • B
    $\frac{-1}{4}$
  • C
    $\frac{1}{2}$
  • D
    $\frac{-1}{2}$

Explore More

Similar Questions

In $\triangle ABC$,if $r=3$ and $R=5$,then $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = $

In $\triangle ABC$,with usual notations,if $\cos \frac{B}{2} = \sqrt{\frac{c+a}{2a}}$,then $a^2 =$

In a $\triangle ABC$,it is known that $AB=AC$. Suppose $D$ is the mid-point of $AC$ and $BD=BC=2$. Then,the area of the $\triangle ABC$ is

If in a triangle $ABC$,$\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$,then $\cos A$ is equal to

In a $\triangle ABC$,if $\angle C = 90^{\circ}$,then $\frac{a^2-b^2}{a^2+b^2}$ is equal to

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