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$△ABC$ - triangle with $45°, 105°$ and $30°.$

Perimeter of triangle is $\sqrt6 + 2\sqrt3 - \sqrt2.$

Find the longest side?

Do you have any ideas? Seems something to do with Law of sines and cosines, something else?

Thank you!

P.S.: how to use Latex?

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    $\begingroup$ what have you tried? Write a few attempts of your idea. $\endgroup$
    – Lion Heart
    Aug 9 at 17:43
  • $\begingroup$ Okay, I started with sines law as x/sin30=y/sin45, so I found the relations between x and y. Then third side via cosine law and x,y. And then with x+y+z=P, and previous equations. But, at least one x or y are superfluous. Like than I tried. $\endgroup$
    – TomJK-st
    Aug 9 at 17:49
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    $\begingroup$ Hi! You should edit your explanation into the original question post, rather than just leaving it in the comments. Comments can disappear at any moment without chance for recovery. (Don't ask me how I know that. <g>) $\endgroup$
    – Brian Tung
    Aug 9 at 17:52

2 Answers 2

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enter image description here Draw $AD\perp BC$

Let $AD=x$ then $AC=2x , DC=x\sqrt3 , BD=x , AB=x\sqrt2$

Perimeter $\triangle ABC = 3x+x\sqrt2+x\sqrt3=\sqrt6 +2\sqrt3-\sqrt2$

the longest side $=x(1+\sqrt3)= (1+ \sqrt 3)\frac{\sqrt6+2\sqrt3-\sqrt2}{3+\sqrt3+\sqrt2}=\frac{6+2\sqrt3+2\sqrt2}{3+\sqrt3+\sqrt2}=2$

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The longest side will be the side that is opposite to the largest angle of the triangle.

You haven't defined the angles clearly so I can't tell which side is largest but the approach remains same.

No need of any law of sines or cosines.

EDIT

Let the sided opposite to $105^o$ be $a$, opposite to $45^o$ be $b$ and opposite to $30^o$ be $c$.

Now by law of sines: $$\frac{a}{\operatorname{sin}105}=\frac{b}{\operatorname{sin}45}=\frac{c}{\operatorname{sin}30}$$ or $$\frac{a}{\frac{1+\sqrt3}{2\sqrt2}}=\frac{b}{\frac{1}{\sqrt2}}=\frac{c}{\frac{1}{2}}$$ or $$b=\frac{2a}{1+\sqrt3} $$ and $$c=\frac{\sqrt2\cdot a}{1+\sqrt3}$$ $\implies$ $$a+\frac{2a}{1+\sqrt3}+\frac{\sqrt2\cdot a}{1+\sqrt3}=\sqrt6+2\sqrt3-\sqrt2$$ After rigorous solving we get, $$a=2$$ Hence the largest side is $2$

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  • $\begingroup$ The longest side will be the side that is opposite to the largest angle of the triangle - CORRECT. $\endgroup$
    – TomJK-st
    Aug 9 at 17:33
  • $\begingroup$ We have only perimeter of whole triangle, without any simple side. Just angles. $\endgroup$
    – TomJK-st
    Aug 9 at 17:34

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