<s>[INST] 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?
 [/INST] $$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{3+1-6}{2\cdot\sqrt 3\cdot 1}=-\frac1{\sqrt 3}$$
Negative sign means $A>90^\circ$. If an angle is greater than $90^\circ$, the other two must be less.
$$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{6+1-3}{2\cdot{\sqrt 6}\cdot 1}=\frac{2}{\sqrt 6}$$
Then $$\sin B=\sqrt{1-\cos^2 B}=\sqrt{1-\frac 46}=\frac 1{\sqrt 3}$$
Using the formula for $$\cos(x+y)=\cos x\cos y-\sin x\sin y$$will get you the result
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