Trigonometry & Inverse Trigonometry
Equilateral triangle, inradius, area, incentre dot product
nta_pyq_2023_jan
Grade 11
Question:
For a triangle ABC, the value of $\cos 2A + \cos 2B + \cos 2C$ is least. If its inradius is 3 and incentre is M, then which of the following is NOT correct?
Perimeter of $\triangle ABC$ is $18\sqrt{3}$
$\sin 2A + \sin 2B + \sin 2C = \sin A + \sin B + \sin C$
$\overrightarrow{MA} \cdot \overrightarrow{MB} = -18$
area of $\triangle ABC$ is $\frac{27\sqrt{3}}{2}$
Step-by-Step Solution
Key Concept: $\cos 2A + \cos 2B + \cos 2C$ is minimized for equilateral triangle ($A=B=C=60°$); use inradius formula $r = a/(2\sqrt{3})$ to find side $a$
Minimum of $\cos 2A+\cos 2B+\cos 2C$ occurs for equilateral triangle. Inradius $r=3$: $r = a/(2\sqrt{3}) \Rightarrow a = 6\sqrt{3}$. Perimeter $= 18\sqrt{3}$ ✓. Area $= \frac{\sqrt{3}}{4}(6\sqrt{3})^2 = \frac{\sqrt{3}}{4}\times 108 = 27\sqrt{3} \neq \frac{27\sqrt{3}}{2}$ — option (4) is NOT correct. $\overrightarrow{MA}\cdot\overrightarrow{MB}$: $|MA|=|MB|= r/\sin(30°)= 3/(1/2)=6$, angle $\angle AMB = 180°-C/2... = 120°$. $\overrightarrow{MA}\cdot\overrightarrow{MB} = 6\times6\times\cos(120°) = -18$ ✓. Answer: (4)
Correct Answer: area of $\triangle ABC$ is $\frac{27\sqrt{3}}{2}$