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QJEEM23Maaee6: JEE Main 2023 Mathematics - Class 11 JEEM23Maaee6 Solution: For a triangle , the value of is least. If its inradius is 3…

For a triangle ABCABC , the value of cos⁡2A+cos⁡2B+cos⁡2C\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?

Multiple Choice Options

  • A MA→ . MB→=−18\overrightarrow {MA} \,.\,\overrightarrow {MB} = - 18
  • B sin⁡2A+sin⁡2B+sin⁡2C=sin⁡A+sin⁡B+sin⁡C\sin 2A + \sin 2B + \sin 2C = \sin A + \sin B + \sin C
  • C perimeter of ΔABC\Delta ABC is 18 3\sqrt3
  • D area of ΔABC\Delta ABC is 2732{{27\sqrt 3 } \over 2}

Step-by-Step Verified Solution

JEE Main 2023 (Online) 1st February Morning Shift Mathematics - Properties of Triangle Question 11 English Explanation

If cos⁡2 A+cos⁡2 B+cos⁡2C\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C} is minimum then A=\mathrm{A}=

B=C=60∘\mathrm{B}=\mathrm{C}=60^{\circ}

So △ABC\triangle \mathrm{ABC} is equilateral

Now in-radias r=3r=3

So in △MBD\triangle \mathrm{MBD} we have

amp;tan⁡30∘=MDBD=ra/2=6aamp;1/3=1a=a=63\begin{aligned}& \operatorname{tan} 30^{\circ}=\frac{M D}{B D}=\frac{r}{a / 2}=\frac{6}{a} \\\\& 1 / \sqrt{3}=\frac{1}{a}=a=6 \sqrt{3}\end{aligned}

Perimeter of △ABC=183\triangle \mathrm{ABC}=18 \sqrt{3}

Area of △ABC=34a2=273\triangle \mathrm{ABC}=\frac{\sqrt{3}}{4} a^2=27 \sqrt{3}

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