For a triangle ABC,the value of cos 2A + cos | vedclass

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

Question

(D) The expression $\cos 2A + \cos 2B + \cos 2C$ is minimized when $	riangle ABC$ is an equilateral triangle,i.e.,$A = B = C = 60^{\circ}$.Given inra

Vedclass · Accepted Answer

Area of $	riangle ABC$ is $\frac{27\sqrt{3}}{2}$

Vedclass · Answer

(D) The expression $\cos 2A + \cos 2B + \cos 2C$ is minimized when $	riangle ABC$ is an equilateral triangle,i.e.,$A = B = C = 60^{\circ}$.Given inradius $r = 3$. For an equilateral triangle,$r = \frac{a}{2\sqrt{3}}$,so $3 = \frac{a}{2\sqrt{3}} \implies a = 6\sqrt{3}$.Perimeter $= 3a = 18\sqrt{3}$. (Option $A$ is correct).Area $= \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (6\sqrt{3})^2 = \frac{\sqrt{3}}{4} (108) = 27\sqrt{3}$. (Option $D$ is incorrect as it states $\frac{27\sqrt{3}}{2}$).For $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$ and $\sin A + \sin B + \sin C = 3\sin 60^{\circ} = \frac{3\sqrt{3}}{2}$. Since $4\sin^3 60^{\circ} = 4(\frac{\sqrt{3}}{2})^3 = \frac{3\sqrt{3}}{2}$,Option $B$ is correct.For $\overrightarrow{MA} \cdot \overrightarrow{MB} = |MA||MB| \cos(120^{\circ}) = (2r)(2r)(-1/2) = -2r^2 = -2(3^2) = -18$. (Option $C$ is correct).

For a triangle $ABC$,the value of $\cos 2A + \cos 2B + \cos 2C$ is least. If its inradius is $3$ and incentre is $M$,then which of the following is $NOT$ correct?

Similar Questions

In a triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $a : b : c =$

In a triangle $ABC$,with usual notations $\angle A=60^{\circ}$,then $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=$

If $A, B, C$ are the angles of a triangle,then $\sum \frac{\cot A + \cot B}{\tan A + \tan B} = $

With usual notation in a $\Delta ABC$,$\left( \frac{1}{r_1} + \frac{1}{r_2} \right) \left( \frac{1}{r_2} + \frac{1}{r_3} \right) \left( \frac{1}{r_3} + \frac{1}{r_1} \right) = \frac{K R^3}{a^2 b^2 c^2}$ where $K$ has the value equal to

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module