Find the longest side of a triangle, given its angles and perimeter $△ABC$ - 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?
 A: 
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 $=x(1+\sqrt3)= (1+ \sqrt 3)\frac{\sqrt6+2\sqrt3-\sqrt2}{3+\sqrt3+\sqrt2}=\frac{6+2\sqrt3+2\sqrt2}{3+\sqrt3+\sqrt2}=2$
A: 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 $2$
