JEE Mains · Maths · STD 11 - Trigonometrical equations
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?
- A Perimeter of \(\triangle ABC\) is \(18 \sqrt{3}\)
- B \(\sin 2 A +\sin 2 B +\sin 2 C =\sin A +\sin B +\sin C\)
- C \(\overrightarrow{ MA } \cdot \overrightarrow{ MB }=-18\)
- D area of \(\triangle ABC\) is \(\frac{27 \sqrt{3}}{2}\)
Answer & Solution
Correct Answer
(D) area of \(\triangle ABC\) is \(\frac{27 \sqrt{3}}{2}\)
Step-by-step Solution
Detailed explanation
If \(\cos 2 A +\cos 2 B +\cos 2 C\) is minimum then \(A =\) \(B = C =60^{\circ}\) So \(\triangle ABC\) is equilateral Now in-radias \(r=3\) So in \(\triangle MBD\) we have \(\operatorname{Tan} 30^{\circ}=\frac{M D}{B D}=\frac{r}{a / 2}=\frac{6}{a}\)…
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