We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$ We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$
I just did the question above in the following way:
$t=\frac{\sqrt{6}+\sqrt{3}+1}{2}$
From Heron we have that:
$=\sqrt{\frac{\sqrt{6}+\sqrt{3}+1}{2}*\frac{\sqrt{3}+1-\sqrt{6}}{2}*\frac{\sqrt{6}-\sqrt{3}+1}{2}*\frac{\sqrt{6}+\sqrt{3}-1}{2}}$
$=\frac{1}{4}\sqrt{8}$
I state that AK is a height of the triangle.
We have that it is enough if $sin(90+B)=sin(A)$
However we also have that $sin(90+B)=sin(BAK)$ and $sin(BAK)=\frac{BK}{1}=BK$ (BAK is a right angled triangle)
Also:
$\frac{1}{2}*sin(A)*\sqrt{3}=\frac{1}{4}\sqrt{8}$
$sin(A)=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}$
So we just have to prove that $BK=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}$
We have that:
$\frac{AK*BC}{2}=\frac{1}{4}\sqrt{8}$
$AK=\frac{\sqrt{4}}{2\sqrt{3}}$
Hence from Pythagoras we have that:
$BK=\sqrt{\frac{8}{12}}=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}=sin(A)$
Hence it is proved.
I realize that my method of proving it is complex. Could you please show some simple solutions to this problem?

Cosine law for $A$ and $B$: $$6=3+1-2\sqrt3\cos A$$ $$3=6+1-2\sqrt6\cos B$$ $$\therefore\quad\cos A=-\frac{1}{\sqrt3},\qquad \cos B=\sqrt{\frac{2}{3}}$$
$$\therefore\quad\cos(A-B)=-\sqrt\frac{1}{3}\sqrt{\frac{2}{3}}+\sqrt{\frac{2}{3}}\sqrt\frac{1}{3}=0$$
