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One angle is $30$ degrees and the opposite side is $10$. Two other sides are $x$ and $12$. What is the maximum possible area of given triangle?

I used law of cosines to find max length of the $x$ then $A = \frac 12\cdot12\cdot x\cdot\sin(30)$ to find the area. The answer I got was around $55.1769$... is this correct?

Hopefully you'll understand my question.

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    $\begingroup$ Did you mean $55.1769$ with a decimal in between? (in place of the comma) $\endgroup$ Jan 30, 2018 at 16:44
  • $\begingroup$ Yes, sorry, we use comma here. $\endgroup$
    – Gueoa
    Jan 30, 2018 at 16:49

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By law of cosines $$100=x^2+144-2\cdot12x\cos30^{\circ}$$ or $$x^2-12\sqrt3x+44=0,$$ which for the maximum of the area gives $$x=6\sqrt3+\sqrt{108-44}$$ 0r $$x=8+6\sqrt3$$ and we obtain: $$S_{\Delta}=\frac{1}{2}\cdot12\cdot(8+6\sqrt3)\cdot\frac{1}{2}=24+18\sqrt3$$

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