If $P$ is a point on the altitude $AD$ of the $\triangle ABC$,and $\angle ABP = \frac{2B}{3}$,then $AP$ is equal to

  • A
    $C \sin \frac{B}{3}$
  • B
    $2C \sin \frac{B}{3}$
  • C
    $C \sin \frac{2B}{3}$
  • D
    $2C \sin \frac{2B}{3}$

Explore More

Similar Questions

In $\triangle ABC$,if $\angle C = \frac{\pi}{3}$,then $\frac{3}{a+b+c} - \frac{1}{a+c}$ equals

The sides of a triangle inscribed in a given circle subtend angles $\alpha, \beta, \gamma$ at the center. The minimum value of the $A.M.$ of $\cos (\alpha + \frac{\pi}{2})$,$\cos (\beta + \frac{\pi}{2})$ and $\cos (\gamma + \frac{\pi}{2})$ is equal to

Match the items of List-$I$ with those of List-$II$ (Here $\Delta$ denotes the area of $\triangle ABC$.)
List-$I$List-$II$
$(A)$ $\sum \cot A$$(i)$ $\frac{(a+b+c)^2}{4\Delta}$
$(B)$ $\sum \cot \frac{A}{2}$$(ii)$ $\frac{a^2+b^2+c^2}{4\Delta}$
$(C)$ If $\tan A : \tan B : \tan C = 1 : 2 : 3$,then $\sin A : \sin B : \sin C =$$(iii)$ $8 : 6 : 5$
$(D)$ If $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$,then $a : b : c =$$(iv)$ $12 : 5 : 13$
$(v)$ $\sqrt{5} : 2\sqrt{2} : 3$
$(vi)$ $4\Delta$

Then the correct match is

In a $\triangle ABC$, $\tan A$ and $\tan B$ are the roots of the equation $pq(x^{2}+1) = r^{2}x$. Then, $\triangle ABC$ is:

The number of triangles in which $\tan A + \tan B + \tan C = \cot A + \cot B + \cot C$ is:

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