1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Did you mean $55.1769$ with a decimal in between? (in place of the comma) $\endgroup$ – Gaurang Tandon Jan 30 '18 at 16:44
  • $\begingroup$ Yes, sorry, we use comma here. $\endgroup$ – Gueoa Jan 30 '18 at 16:49
1
$\begingroup$

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: $$S_{\Delta}=\frac{1}{2}\cdot12\cdot(8+6\sqrt3)\cdot\frac{1}{2}=24+18\sqrt3$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.