Limit of an area I have to find the area coloured in grey when $\beta\to (\pi/2)^-$. ^-$, then the other angle goes to zero, so the grey area goes to zero. For example if $\beta=89^{\circ}$ then the triangle looks like this ![enter image description here]( but I don't know if this is a rigorous way to prove that the grey area is zero. any help is appreciated. thank you.

First note that $\tan \beta=H/10 \implies H=10 \tan \beta$.

From this we can deduce that the area of the whole triangle is $50 \tan \beta $, by using the usual area triangle.

What is the area of the arc in terms of $\beta$?* Let's call it $A_{\beta}$. Then $A_{grey}=50\tan \beta-A_{\beta}$.

Now, all you have to do is evaluate $$\lim_{\beta \to \pi/2} 50 \tan \beta-A_{\beta}.$$

*: as a hint, it is some fraction of a circle of radius $10$.