Hello, fxsapa!

We have: point O, anywhere on the plane; line AB, anywhere on the plane.

Point M, moving on AB from A, at speed described by function s(t).

Find the rate of change of $\displaystyle \angle AOM$ (from time t),

given that you can know everything about $\displaystyle \Delta ABO$ (lengths, angles, ...)

but that this triangle is scalene. Code:

O
*
*/ *
* / *
b * θ/ *
* / *
* / *
* / *
* φ / *
A * * * * * * * * * * * B
x M

We have: .$\displaystyle \Delta AOB,\:A = \angle OAB,\;\theta = \angle AOM,\;b = OA,\;x = AM$

. . . .and: .$\displaystyle \phi = \angle AMO \:=\:180^o - A - \theta$

In $\displaystyle \Delta OAM$, apply the Law of Sines:

. . $\displaystyle \frac{\sin\theta}{x} \:=\:\frac{\sin(180-A-\theta)}{b} \quad\Rightarrow\quad b\sin\theta \;=\;x\sin([180 - A] - \theta)

$

. . $\displaystyle b\sin\theta \;=\;x\bigg[\underbrace{\sin(180-A)}_{\text{This is }\sin A}\cos\theta - \underbrace{\cos(180-A)}_{\text{This is }-\cos A}\sin\theta\bigg] $

. . $\displaystyle b\sin\theta \;=\;x\underbrace{\bigg[\sin A\cos\theta + \cos A\sin\theta\bigg]}_{\text{This is }\sin(\theta + A)} $

. . $\displaystyle b\sin\theta \;=\;x\sin(\theta + A)$

Differentiate with respect to time:

. . $\displaystyle b\cos\theta\,\frac{d\theta}{dt} \;=\; x\cos(\theta + A)\,\frac{d\theta}{dt} + \sin(\theta + A)\,\frac{dx}{dt}$

. . $\displaystyle b\cos\theta\,\frac{d\theta}{dt} - x\cos(\theta + A)\,\frac{d\theta}{dt} \;=\;\sin(\theta + A)\,\frac{dx}{dt}$

. . $\displaystyle \bigg[b\cos\theta - x\cos(\theta+A)\bigg]\,\frac{d\theta}{dt} \;=\;\sin(\theta + A)\,\frac{dx}{dt}$

Therefore: .$\displaystyle \frac{d\theta}{dt} \;=\;\frac{\sin(\theta+A)\,\frac{dx}{dt}}{b\cos\th eta - x\cos(\theta +A)} $