If in a triangle ABC,cos A + cos B + cos C = | vedclass

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

Question

(B) We know that for any triangle $ABC$,the maximum value of $\cos A + \cos B + \cos C$ is $\frac{3}{2}$,which occurs if and only if $A = B = C = 60^{

Vedclass · Accepted Answer

Equilateral

Vedclass · Answer

(B) We know that for any triangle $ABC$,the maximum value of $\cos A + \cos B + \cos C$ is $\frac{3}{2}$,which occurs if and only if $A = B = C = 60^{\circ}$.Alternatively,using the identity $\cos A + \cos B + \cos C = 1 + 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{3}{2}$,we get $\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{1}{8}$.Since the maximum value of $\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$ is $\frac{1}{8}$ (by Jensen's inequality for concave functions),this equality holds only when $\frac{A}{2} = \frac{B}{2} = \frac{C}{2} = 30^{\circ}$,i.e.,$A = B = C = 60^{\circ}$.Thus,the triangle is equilateral.

If in a triangle $ABC$,$\cos A + \cos B + \cos C = \frac{3}{2}$,then the triangle is

Similar Questions

In a triangle $ABC$,with usual notations,if $3b = a + c$,then $\cot \frac{A}{2} \cdot \cot \frac{C}{2} = $

In a triangle $ABC$,with usual notations $a=2, b=3, c=5$,then $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=$

In a $\Delta ABC,$ if $A:B:C = 3:5:4,$ then $a + b + c\sqrt{2}$ is equal to

If in a triangle $ABC$,$\frac{\sin A}{4} = \frac{\sin B}{5} = \frac{\sin C}{6}$,then the value of $\cos A + \cos B + \cos C$ is equal to

In a $\Delta ABC$,let $a, b$ and $c$ denote the lengths of sides opposite to vertices $A, B$ and $C$ respectively. If $b = 2, c = \sqrt{3}$ and $\angle BAC = \frac{\pi}{6}$,then the value of the circumradius of triangle $ABC$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module