If $p_1, p_2, p_3$ are the altitudes and $a=4, b=5, c=6$ are the sides of a triangle $ABC$,then $\frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} =$

  • A
    $\frac{77}{225}$
  • B
    $\frac{44}{225}$
  • C
    $\frac{308}{225}$
  • D
    $\frac{22}{75}$

Explore More

Similar Questions

If $\alpha, \beta$ and $\gamma$ are the lengths of the altitudes of a $\triangle ABC$ with area $\Delta$,then $\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$ is equal to

If $ABCD$ is a cyclic quadrilateral,then the value of $\cos A - \cos B + \cos C - \cos D = $

In a triangle $ABC$,with usual notations,the sides $a, b, c$ are the roots of the equation $x^3-11x^2+38x-40=0$. Then,find the value of $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$.

In any $\triangle ABC$,$b^2 \sin 2C + c^2 \sin 2B$ is equal to

Consider a triangle $PQR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$,respectively. Then which of the following statements is (are) $TRUE$?
$(A)$ $\cos P \geq 1-\frac{p^2}{2qr}$
$(B)$ $\cos R \geq \left(\frac{q-r}{p+q}\right) \cos P + \left(\frac{p-r}{p+q}\right) \cos Q$
$(C)$ $\frac{q+r}{p} < 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}$
$(D)$ If $p < q$ and $p < r$,then $\cos Q > \frac{p}{r}$ and $\cos R > \frac{p}{q}$

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