We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$

Question

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?

Answer 1

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

$$\therefore\quad\cos(A-B)=-\sqrt\frac{1}{3}\sqrt{\frac{2}{3}}+\sqrt{\frac{2}{3}}\sqrt\frac{1}{3}=0$$

Answer 2

We are to make use of $$ \cos (90+\theta) = -\sin \theta$$

Let $a=\sqrt{6}$, $b=\sqrt{3}$, $c=1$. So we assumed $a>b>c$ or, $A>B>C$.

Now use cosine-law,

$$ \cos A = \dfrac{b^2+c^2-a^2}{2bc}$$

and show by calculation that

$$ \cos A = -\sin B $$

where, $\sin B = \sqrt{1-\cos^2 B}$.

Answer 3

$$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{3+1-6}{2\cdot\sqrt 3\cdot 1}=-\frac1{\sqrt 3}$$ 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