three altitude intersecting each other at H.
Find the area of triangle ABC
My attempt feel like cheating a little bit, I let BA = $\dot{\vec{C}}$, AC = $\dot{\vec{B}}$ and CB = $\dot{\vec{A}}$
then I let HB = $\dot{\vec{X}}$ , HC= $\dot{\vec{Y}}$ and HA =$\dot{\vec{Z}}$
i have that $\dot{\vec{A}+\vec{B}+\vec{C}=0}$
$\dot{(\vec{X}+\vec{Y})+(\vec{Y}+\vec{Z})+(\vec{Z}+\vec{X})=0}$
$\dot{\vec{X}+\vec{Y}+\vec{Z}=0}$
I let $\dot{\vec{X}}$= $\sqrt{3}-1j$
$\dot{\vec{Y}}$= $ai+bj$
$\dot{\vec{Z}}$= $-ai+cj$
I know the absolute of each vector so $a^2+b^2=4$ and $a^2+c^2=2$ , thus $b^2-c^2 = 2 $
I know that $b+c=1-\sqrt{3}$ , this mean that $b-c= -\sqrt{3} -1 $
I have $b= \sqrt{3}$, $c=1$ and $a= 1$
then area of triangle $ABC = \frac{1}{2}(\left |X\times Y\right |+\left |Y\times Z\right |+\left |Z\times X\right |)$ = $\frac{3\sqrt{3}-1}{2}$
I'm not sure if this is correct the area seem too small and, from the look of the question, I'm not supposed to solve this with vector because they give me rays of altitude and i think that i'm supposed to use that somehow knowing some properties
Do you guys have any other ways of solving this not using vector? Thanks.
P.S. my attempt is wrong because i cant add vector like that lol.
So you know $ABC$ is acute-angled (since $H$ is inside $\triangle ABC$), and we have \begin{align*} 2R\cos A&=AH=\sqrt2,\\ 2R\cos B&=BH=\sqrt3-1,\\ 2R\cos C&=CH=2,\\ A+B+C&=\pi. \end{align*} Solving, you get $R=\sqrt 2$, $A=\pi/3$, $C=\pi/4$, $B=5\pi/12$. Hence
$[ABC]=2R^2\sin A\sin B\sin C=\sqrt{3+\sqrt3}/2\approx 2.366$