For a triangle ABC, 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? [2023]

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Solution

Q.
For a triangle ABC, 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 [2023]

1 area of $∆ A B C$ is $\frac{27 \sqrt{3}}{2}$

2 $\sin 2 A + \sin 2 B + \sin 2 C = \sin A + \sin B + \sin C$

3 perimeter of $∆ A B C$ is $18 \sqrt{3}$

4 $\vec{M A} \cdot \vec{M B} = - 18$

Ans.
(1)

$Here, \cos 2 A + \cos 2 B + \cos 2 C = - \frac{3}{2}$

$\Rightarrow 2 A = 2 B = 2 C = 120 ° \Rightarrow A = B = C = 60 °$

$Now, inradius = \frac{Area of ∆ A B C}{Semiperimeter}$

$3 = \frac{\frac{\sqrt{3}}{4} a^{2}}{\frac{3 a}{2}}$ $[Where a = side of ∆]$

$\Rightarrow a = \frac{18}{\sqrt{3}} = 6 \sqrt{3}$

$Perimeter of ∆ A B C = 18 \sqrt{3}$

$Area of ∆ A B C = \frac{\sqrt{3}}{4} \times {(6 \sqrt{3})}^{2} = 27 \sqrt{3}$

$A = 60 \Rightarrow 2 A = 120$

$\sin 2 A = \sin A$

$\sum \sin 2 A = \sum \sin A$

$| \vec{M A} |$ $=$ $| \vec{M B} |$ $\Rightarrow 6 = 2 r$

$\vec{M A} \cdot \vec{M B} = 36 \cos 120 ° = - 18$

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