What is the value of $\Delta\gamma/\Delta\theta$ in the attached figure?

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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 $\triangle AZY$.

Answer 4

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.