How to calculate green area plus red area without using integral calculus? I found this problem on the net, and the truth is I don't know where to start, someone would be so kind to explain it to me, I would be very grateful.
edited

 A: Expanding-upon a comment ...

Taking $A$ and $B$ as endpoints of segments $b$ and $a$ as shown, define $\alpha$ and $\beta$ the inscribed angles angles at this points, subtending chords $a$ and $b$. Let the cross-chords meet at $P$, making an angle of $\theta$ there; the problem specifically has $\theta=60^\circ$, but I like to work in generalities whenever possible, to see how parameters contribute to the outcome. (Note that, by a cousin of the Inscribed Angle Theorem, we know that $\theta=\alpha+\beta$.) Let the circle have radius $r$.

Naming the "other" endpoint of chord $a$ as $P'$, we can find $A'$ so that $\triangle PAB\sim\triangle P'A'B$. As $\overline{P'A'}$ subtends an angle $\beta$, it must have length $b$.
Now, by the (Extended) Law of Sines and the Law of Cosines, we can find the length of $\overline{A'B}$ in two ways:
$$(2r\sin P')^2=|A'B|^2=a^2+b^2-2a b\cos P'
\quad\to\quad r^2=\frac{a^2+b^2+2ab\cos\theta}{4\sin^2\theta}$$
Thus, radius $r$ is determined by $a$, $b$, and $\theta$.
Since the circular segment determined by the new $b$ chord matches that of the original, we can leverage it to calculate the target area. Re-focusing ...

$$\begin{align}
\text{target area} &= |\operatorname{sector}BP'A|-|\square OA'P'B| \\[6pt]
&=\frac12r^2\angle A'OP - |\triangle A'OP| - |\triangle A'P'B| \\[6pt]
&=r^2\theta - \frac12r^2\sin2\theta - \frac12ab\sin\theta \\[6pt]
&=r^2\theta - \frac12\left(\frac{a^2+b^2+2ab\cos\theta}{4\sin^2\theta}\cdot 2\sin\theta\cos\theta + ab\sin\theta\right) \\[6pt]
&=r^2\theta - \frac1{4\sin\theta}\left((a^2+b^2)\cos\theta+2ab\cos^2\theta + 2ab\sin^2\theta\right) \\[6pt]
&=r^2\theta - \frac1{4\sin\theta}\left((a^2+b^2)\cos\theta+2ab\right)
\end{align}$$
Thus, the target area itself is expressed in terms of $a$, $b$, and $\theta$, as desired. $\square$
(Considering the final form of the quadrilateral area, I suspect there may be a cleaner path to it.)
A: Here is a solution to the question involving extended sine law and circle angle properties.



Edit: you don't need to find theta for this question, can use trig identities instead to get an exact answer
A: 
There is nothing wrong with the methodology used in kyary’s answer. However, it has generated a surprisingly long thread of comments, mostly exchanges of openions between the asker and the answerer. In view of that, we decided to post the following answer, which gives the complete solution to the problem at hand. We hope that our answer would help OP to clear doubts, if he or she still has some. Please note that certain elements of the method given below are somehow slightly different from that of kyary’s solution.
Let the radius of the circle be $r$. We denote the measure of angles $B\hat{O}A$ and $D\hat{O}C$ by $2\omega$ and $2\phi$ respectively. From this, it follows that
$$ \measuredangle BDA = \omega \quad\text{and}\quad \measuredangle DAC =\phi.$$
The external angle at the vertex $E$ of $\triangle AED$ is equal to $60^o$. Therefore,
$$\phi+\omega=60^o. \tag{1}$$
As shown in the $\mathrm{Fig.\space 1}$, we extend the given configuration by drawing the four radii $OA$, $OB$, $OC$, and $OD$ and the line segment $AD$. Addition of the radii gives us two isosceles triangles $BOA$ and $DOC$, using which we shall write,
$$a=2r\sin \left(\phi\right) \quad\text{and}\quad b=2r\sin \left(\omega\right).$$
When we remove $r$ from these two equations, we get
$$b sin \left(\phi\right) = a\sin \left(\omega\right).$$
Therefore, we have
$$ b sin \left(\phi\right) = a\sin \left(60^o - \phi\right) = \frac{a}{2} \left(\sqrt{3}\cos \left(\phi\right) -\sin \left(\phi\right)\right).$$
By simplifying this trigonometric identity as illustrated below, we can obtain a value for $\phi$ in terms of $a$ and $b$.
$$\phi = \tan^{-1} \left(\frac{\sqrt{3}a}{a+2b}\right)$$
There exists a very similar expression for $\omega$ as well.
$$\omega = \tan^{-1} \left(\frac{\sqrt{3}b}{b+2a}\right)$$
Now, we are in a position to express the radius $r$ in terms of $a$ and $b$.
$$r=\frac{a}{2\sin \left(\phi\right)}=\frac{a}{2\sin \left[\tan^{-1} \left(\dfrac{\sqrt{3}a}{a+2b}\right)\right]} \tag{2}$$
By the way, we need an expression for $\left(\phi - \omega\right)$ as well. We assume that you are versed in manipulating inverse trigonometric functions.
$$\phi - \omega =  \tan^{-1} \left(\frac{\sqrt{3}a}{a+2b}\right) - \tan^{-1} \left(\frac{\sqrt{3}b}{b+2a}\right) = \tan^{-1}\left(\frac{\dfrac{\sqrt{3}a}{a+2b}-\dfrac{\sqrt{3}b}{b+2a}}{1+ \dfrac{\sqrt{3}a}{a+2b}\cdot \dfrac{\sqrt{3}b}{b+2a}}\right)$$
Upon simplifying, the above expression yields the following equation.
$$\phi - \omega = \tan^{-1}\left(\frac{\sqrt{3}\left(a^2 – b^2\right)}{ a^2 + b^2+4ab}\right) \tag{3}$$
Now, we can express the two areas $A_{a}$ and $A_{b}$.
$$A_{a}=r^2\phi – \frac{1}{2}r^2\sin \left(2\phi\right)$$
$$A_{b}=r^2\omega– \frac{1}{2}r^2\sin \left(2\omega\right)$$
$$\therefore\quad A_{a} + A_{b}=\left[\phi + \omega - \frac{1}{2}\Big(\sin \left(2\phi\right) + \sin \left(2\omega\right)\Big)\right]r^2$$
$$\qquad\qquad\qquad = \Big[\phi + \omega - \sin\left(\phi + \omega\right) \cos\left(\phi - \omega\right) \Big]r^2. \tag{4}$$
By substituting values from (1), (2), and (3) in (4), we obtain,
$$ A_{a} + A_{b} =\frac{2\pi-3\sqrt{3}\cos\left[\tan^{-1}\left(\dfrac{\sqrt{3}\left(a^2 – b^2\right)}{ a^2 + b^2+4ab}\right) \right]}{24\sin^2\left[\tan^{-1}\left(\dfrac{\sqrt{3}a}{a+2b}\right)\right]}a^2$$
