Change of angle given a load (geometric problem)

Question

I have the following figure, where a load $P$ is applied to this unit, which causes the angle $\theta$ as well as $u$ to increase. The sides AC,BC etc. are rigid and have length $L$, while AB is $2l$ which is elastic (hence $sin\theta = \frac{l}{L}$).

I have to show: $\Delta l=Lcos(\theta)\Delta\theta$, but how can I do this? I know it makes sense but I do not have an idea how I can use the $\Delta\theta$ in this problem. Perhaps I can use that when a load $P$ is applied, the angle $\Delta\theta$ and length $\Delta l$ increase, so $P=\frac{\Delta l}{\Delta \theta}$, and than using $cos\theta = \frac{P}{L}$ and the expression for $sin\theta$. However I think this is too short and I can give a better explanation, so any help/hints would be appreciated!

Answer 1

(1) Put $L$ in other position so $P$ becomes $P'$ and you get $l+\Delta l$ in the horizontal line.

(2) Draw a line parallel to $CB$ passing through $P'$ so you have the angle $\theta+\Delta \theta$.

$\sin(\theta+\Delta\theta)=\dfrac{l+\Delta l}{L}=\sin\theta\cos \Delta \theta+\cos\theta\sin\Delta\theta=\dfrac lL\cos\Delta \theta+\cos\theta\sin\Delta\theta$

You can do $\cos\Delta\theta=1$ because you are working with differentials (this is the only justification to your formula that I find. Keep in mind that the equation I have written is true). Thus you get your formula.

Answer 2

Suppose $\theta$ is increased by $d\theta$. If point C is fixed then B moves a distance $Ld\theta$ perpendicular to CB. The horizontal projection of this distance is $Ld\theta\cos\theta$.