What is the value of $\Delta\gamma/\Delta\theta$ in the attached figure? Well, I encountered a problem, and then simplified it and put here, however according to the first answer, I found out that I made a mistake during the simplifying the problem.
Here is the original problem:
As you may see in the attached figure, we have a triangle in which the lengths of the sides $L_1$ and $L_3$ are constant. Also, the vertices $O$ and $a$ are fixed. 
$\theta$ is the angle between the horizontal axis and $L_3$ (see the figure).
$\gamma$ is the angle between the horizontal axis and $L_2$ (see the figure).
Now, the question is:
When the vertex $b$ moves such that all the assumptions will be held, what would be the proportion of $\Delta\gamma$ over $\Delta\theta$. 
I think that the vertex $b$ can move on a circle of radius $L_3$ and center of $O$. I appreciate any help in advance.

 A: The picture clearly show $\dfrac {\beta}{\alpha}$ is not constant.

A: Here is a complete solution using $\tan$ and $\tan^{-1}$.
Let us fix notations, with points 
$a=(x_a,y_a)=(L_1\cos(\theta_1),L_1\sin(\theta_1))$ and
$b=(L_3 \cos(\theta), L_3 \sin(\theta))$. 
We have:
$$\tag{1}\vec{ba}\binom{x_a-L_3 \cos(\theta)}{y_a-L_3 \sin(\theta)}.$$ 
It is not difficult to see that the (negative) angle $-\alpha$ that $\vec{ba}$ makes with respect to the horizontal reference is such that $\alpha+\gamma=\pi$. Thus, using (1):
$$\tan(\alpha)=-\left(\frac{y_a-L_3 \sin(\theta)}{x_a-L_3 \cos(\theta)}\right)$$
Therefore, 
$$\tag{2}\gamma=\pi-\alpha=\pi+\tan^{-1}\left(\frac{y_a-L_3 \sin(\theta)}{x_a-L_3 \cos(\theta)}\right).$$
It suffices now to differentiate (2) (I used Mathematica for that) to obtain:

$$\tag{3}\dfrac{\Delta \gamma}{\Delta \theta}\approx\dfrac{E-L_3^2}{L_1^2+L_3^2-2E} \ \ \text{with} \ \ E:=L_3(x_a\cos(\theta)+y_a\sin(\theta))$$

Edit: Formula (3) can be written in a different way:

$$ \dfrac{\Delta \gamma}{\Delta \theta}\approx \dfrac{L_1L_3\cos(\theta_1-\theta)-L_3^2}{L_2^2}$$

Explanation: one can apply  the cosine rule to the denominator of (3): $L_1^2+L_3^2-2L_1 L_3\cos(\theta_1-\theta)=L_2^2$ applied to triangle Oab.
A: Since A is fixed, I assume that it is located at (h, k) where h and k are known quantities.

Further assumptions:- OY = m, OA = n, $\angle XOY = \alpha$. Let the horizontal line through Y cut OA at Z and Z divide OA in the ratio $\rho : 1$. The target is to find $\beta’$ (instead of $\beta$) in term of other known quantities.
$Y = (m \cos \alpha, m \sin \alpha)$.
$Z = (?,m \sin \alpha)$; where  $m \sin \alpha = \dfrac {0 \times 1 + k \times \rho}{\rho + 1}$.
From which, we get $\rho = \dfrac {m \sin \alpha}{k – m \sin \alpha}$, which is then known.
Since Z divides OA (whose length is n) in the ratio $\rho : 1$, $OZ = … = \dfrac {\rho \times n}{\rho + 1}$ and $ZA = … = \dfrac {n}{\rho + 1}$. Also $\theta = \sin^{-1} (\dfrac { m \sin \alpha}{\rho}).$
$\beta’$ can then be found by applying sine/cosine laws to the $\triangle AZY$.
A: I retain the relationship $\alpha = \pi - \gamma$ as proposed by JeanMarie; allow me to explain the coordinates of $a$ using 1 variable only: 
Let slope of $L_1 = \tan \theta_0$. We have
$$
\begin{align}
a&= (L_1 \cos \theta_0, L_1 \sin\theta_0) \\
b&= (L_3 \cos \theta, L_3 \sin\theta)
\end{align}
$$
Finding the slope of $ab$,
$
\tan \alpha = \frac{ L_3 \sin\theta - L_1 \sin\theta_0
}{ L_3 \cos\theta - L_1 \cos\theta_0
}
$ -- (*)
Hence
$
\cos^2 \alpha =
\frac{(L_3 \cos\theta - L_1 \cos\theta_0)^2
}{(L_3 \sin\theta - L_1 \sin\theta_0
)^2+(L_3 \cos\theta - L_1 \cos\theta_0
)^2
}
= \frac{ (L_3 \cos\theta - L_1 \cos\theta_0)^2
}{ L_1^2 + L_3^2 - 2L_1 L_3 \cos(\theta_0-\theta)
}
$
Differentiate both sides of (*),
$
\sec^2 \alpha \text d \alpha = \frac{
L_3^2 - L_1L_3 \cos(\theta_0-\theta)
}{(L_3 \cos\theta - L_1 \cos\theta_0)^2
} \text d \theta
$
$
\frac{\text d \alpha
}{\text d \theta
} =
 \frac{
L_3^2 - L_1L_3 \cos(\theta_0-\theta)
}{(L_3 \cos\theta - L_1 \cos\theta_0)^2
} \cos^2 \alpha
= \frac{
L_3^2 - L_1L_3 \cos(\theta_0-\theta)
}{(L_3 \cos\theta - L_1 \cos\theta_0)^2
} \frac{ (L_3 \cos\theta - L_1 \cos\theta_0)^2
}{ L_1^2 + L_3^2 - 2L_1 L_3 \cos(\theta_0-\theta)
}
= \frac{L_3^2 - L_1L_3 \cos(\theta_0-\theta)
}{L_1^2 + L_3^2 - 2L_1 L_3 \cos(\theta_0-\theta)
}
$
So from $\alpha = \pi - \gamma$, 
$\frac{\text d \gamma
}{\text d \theta
} =-
\frac{\text d \alpha
}{\text d \theta
} 
$
Hence
$$\frac{\text d \gamma
}{\text d \theta
}
= \frac{L_1L_3 \cos(\theta_0-\theta)-L_3^2
}{L_1^2 + L_3^2 - 2L_1 L_3 \cos(\theta_0-\theta)
}
$$
You can see the value as per asked in the question varies with the size of $\theta$, so there is no definite value. 
Edit
By cosine law,
$L_1^2 + L_3^2 - 2L_1 L_3 \cos(\theta_0-\theta) = L_2^2$
I am reluctant in putting $L_2$ in the expression as it is yet another variable. 
My objective was to show that $\text d \gamma / \text d \theta$ is not a constant. 
Introducing $L_2$ might be confusing as it appears to be a constant like $L_1$ or $L_3$, which isn't in reality.
