Suppose that the center of the circles did not lie on one of the altitudes of $\triangle ABC$ (as shown in gray). Then by leaving the base of that altitude alone and moving that altitude to the center, we increase the altitude, and thereby the area of the triangle (as shown in black).
$\hspace{3.5cm}$
Therefore, the orthocenter of the triangle must coincide with the center of the circles.
Consider the following diagram, in which $O$ is both the orthocenter of $\triangle ABC$ and the center of the circles:
$\hspace{3.5cm}$
$\triangle AOF$ is similar to $\triangle COD$, thus, $|\overline{OC}||\overline{OF}|=|\overline{OA}||\overline{OD}|$. Furthermore, $\triangle AOE$ is similar to $\triangle BOD$, thus, $|\overline{OA}||\overline{OD}|=|\overline{OB}||\overline{OE}|$. Therefore, define
$$
p=|\overline{OC}||\overline{OF}|=|\overline{OA}||\overline{OD}|=|\overline{OB}||\overline{OE}|\tag{1}
$$
and
$$
a=|\overline{OA}|\quad b=|\overline{OB}|\quad c=|\overline{OC}|\tag{2}
$$
With these definitions, we get
$$
|\overline{OD}|=p/a\quad|\overline{OE}|=p/b\quad|\overline{OF}|=p/c\tag{3}
$$
Note that
$$
\frac{|\triangle AOB|}{|\triangle ABC|}=\frac{|\overline{OF}|}{|\overline{OF}|+|\overline{OC}|}=\frac{p}{p+c^2}\tag{4}
$$
$$
\frac{|\triangle BOC|}{|\triangle ABC|}=\frac{|\overline{OD}|}{|\overline{OD}|+|\overline{OA}|}=\frac{p}{p+a^2}\tag{5}
$$
$$
\frac{|\triangle COA|}{|\triangle ABC|}=\frac{|\overline{OE}|}{|\overline{OE}|+|\overline{OB}|}=\frac{p}{p+b^2}\tag{6}
$$
and because $|\triangle AOB|+|\triangle BOC|+|\triangle COA|=|\triangle ABC|$, $(4)-(6)$ yield
$$
\frac{p}{p+a^2}+\frac{p}{p+b^2}+\frac{p}{p+c^2}=1\tag{7}
$$
Using $a=1$, $b=2$, and $c=3$, we can solve $(7)$ to get $p=1.458757077431284$.
Looking at the area of $\triangle AOB$ in two different ways, we get
$$
\begin{align}
4|\triangle AOB|^2=|\overline{AB}|^2|\overline{OF}|^2&=(a^2+b^2-2ab\cos(\angle AOB))(p/c)^2\\
&=a^2b^2\sin^2(\angle AOB)\\
&=a^2b^2(1-\cos^2(\angle AOB))\tag{8}
\end{align}
$$
Solving $(8)$ for $\cos(\angle AOB)$ using the quadratic formula, and similarly for the other angles, yields
$$
\cos(\angle AOB)=\frac{p^2-\sqrt{(p^2-c^2a^2)(p^2-b^2c^2)}}{abc^2}\tag{9}
$$
$$
\cos(\angle BOC)=\frac{p^2-\sqrt{(p^2-a^2b^2)(p^2-c^2a^2)}}{a^2bc}\tag{10}
$$
$$
\cos(\angle COA)=\frac{p^2-\sqrt{(p^2-a^2b^2)(p^2-b^2c^2)}}{ab^2c}\tag{11}
$$
Using the equations above, we get
$$\angle BOC=104.071123766006501^\circ$$
$$|\triangle BOC|=2.909984011512956$$
$$\angle COA=119.094556197774592^\circ$$
$$|\triangle COA|=1.310727640381874$$
$$\angle AOB=136.834320036218908^\circ$$
$$|\triangle AOB|=0.684110332666474$$
Therefore, we get
$$|\triangle ABC|=4.904821984561304$$
How do you know $\frac{d}{d\theta}\sin(\theta) = \cos(\theta)$? Is it still true if you measure $\theta$ in degrees instead of radians?
It's actually a remarkably subtle point, and I've seen many good mathematicians get confused by it.
I'll leave out the details of proving the derivative formula for now, but the crux of the matter is that you need to know that $\sin(\theta)$ is very close to $\theta$ when $\theta$ is small, and this is only true if you measure $\theta$ in radians. The best explanation I've seen for that is actually here on math exchange: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
An important implicit point there is that the area of a wedge of a circle $\frac{\theta}{2} r^2$, which only holds if $\theta$ is measured in radians.
Best Answer
You need to express angles in radians, i.e. $2^{\circ} =2 \pi/180$ radians.