Question

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

• A. area of $\triangle A B C$ is $\frac{27 \sqrt{3}}{2}$
• B. $\sin 2 A+\sin 2 B+\sin 2 C=\sin A+\sin B+\sin C$
• C. perimeter of $\triangle A B C$ is $18 \sqrt{3}$
• D. $\overrightarrow{M A} \cdot \overrightarrow{M B}=-18$

Answer 1

The Correct Option is A

If cos2A+cos2B+cos2C is minimum then A= B=C=60∘ 
So △ABC is equilateral 
Now in-radias r=3 
So in △MBD we have 
Tan30∘=BDMD​=a/2r​=a6​ 
1/3​=a1​=a=63​ 
Perimeter of △ABC=183​ 
Area of △ABC=43​​a2=273​

Answer 2

If \( \cos 2A + \cos 2B + \cos 2C \) is minimum, then \[ A = 60^\circ, \quad B = C = 60^\circ \] \[ \Rightarrow \triangle ABC \text{ is equilateral} \] \[ \text{In-radius} = r = 3 \] 

 \[ \text{Area} = \frac{\sqrt{3}}{4} a^2 = 27\sqrt{3} \] \[ \text{Thus, area is incorrectly given as } \frac{27\sqrt{3}}{2} \]