Maximum area of the triangle

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Maximum area of the triangle

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#1 h Jan 19, 2017, 11:42 AM • 1 Y

Y by Adventure10

If in some triangle $\triangle ABC$ we are given :
$\sqrt{3} \cdot \sin(C)=\frac{2- \sin A}{\cos A}$ and one side length of the triangle equals $2$ , then under these conditions find the maximum area of the triangle $ABC$ .

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#2 h Apr 13, 2025, 11:57 AM

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Let us take $\sin(\angle{C}) = \frac{2-\sin(\angle{A})}{\sqrt{3}\cos(\angle{A})},$ and since $\angle{C} \in (0, \pi) \Rightarrow \sin(\angle{C}) \in (0,1]$ we require:

$0 < \frac{2-\sin(\angle{A})}{\sqrt{3}\cos(\angle{A})} \le 1$ ;

or $2-\sin(\angle{A}) \le \sqrt{3}\cos(\angle{A})$ ;

or $4-4\sin(\angle{A})+\sin^{2}(\angle{A}) \le 3[1-\sin^{2}(\angle{A})]$ ;

or $4\sin^{2}(\angle{A}) -4\sin(\angle{A})+1 \le 0$ ;

or $[2\sin(\angle{A})-1]^2 \le 0$ ;

or $\angle{A} = \frac{\pi}{6}, \frac{5\pi}{6}$ (i).

Checking both values of (i) for $\sin(\angle{C}) = \frac{2-\sin(\angle{A})}{\sqrt{3}\cos(\angle{A})}$ yields $\pm 1$ , or $\angle{C} = \frac{\pi}{2}, \frac{3\pi}{2}$ respectively, which only the former angle is admissible $\Rightarrow \angle{A} = \frac{\pi}{6}, \angle{B}=\frac{\pi}{3}, \angle{C}=\frac{\pi}{2}$ . If one of the sides of $\Delta ABC$ has length $2$ , then we have the following triplets $(a,b,c) = (t,\sqrt{3}t,2t)$ :

$(a,b,c) = (2,2\sqrt{3},4); (1,\sqrt{3},2); \left(\frac{2}{\sqrt{3}},2,\frac{4}{\sqrt{3}}\right)$ (ii).

If $A=\frac{1}{2}ab$ is the area, then (ii) yields: $\frac{\sqrt{3}}{2} < \frac{2\sqrt{3}}{3} < 2\sqrt{3} \Rightarrow \textcolor{red}{A_{MAX} = 2\sqrt{3}}$ .

This post has been edited 3 times. Last edited by Mathzeus1024, Apr 13, 2025, 12:02 PM

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