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In my case, we consider an helicoid $\varphi(u,v)= (u \; \sin\, v, u \; \cos\, v,v).$ Its first fundamental form is $$\mathit{I} = \begin{pmatrix} E & F\\ F & G \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & u^2 + 1 \end{pmatrix}.$$

Now, we have to compute the area, the length of the sides and the angles of the triangle, defined by $$0< u< \sinh\, v,\; 0< v< a.$$

For the area I used the formula $A = \int_{0}^{a}\int_{0}^{\sinh\, v}\sqrt{EG-F^2}dudv$. For the sides I have done the following parametrizations $$\alpha _1(t) = (0,t)\quad \text{ where }\quad 0< t< a$$ $$\quad\quad\alpha _2(t) = (t,a)\quad \text{ where }\quad 0< t< \sinh\, a$$ $$\quad\quad\alpha _3(t) = (\sinh\, t,t)\quad \text{ where }\quad 0< t< a.$$ For the length of the curves I have that if $\alpha (t)=(u(t),v(t))$ then $$L = \int_{a}^{b} \sqrt{E(u')^2+2Fu'v'+G(v')^2}dt.$$ And finally, if $\alpha$ and $\beta$ are two curves that intersect in the surface they form an angle $\theta$ such that $$\cos\, \theta=\frac{\mathit{I(\alpha',\beta')}}{\left | \alpha' \right |\left | \beta' \right |}.$$ The problem I have is that the $G$ has an $u^2$ and the lenght of the curves and their angles depends of $u$. So there is something I'm not understanding well. I appreciate any help. Thanks!

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You've set this up well, except for one issue. All you have to do is realize where the vertices of the triangle are (in $(u,v)$ coordinates). One is at $(0,0)$, one is at $(0,a)$, and the last is at $(\sinh a,a)$. So are your curves the correct boundary curves of the triangle? (Hint: One is wrong.) But once you know where the vertices are, there's no problem computing the first fundamental form at this points to find angles.

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  • $\begingroup$ $\alpha_2(t) = (t,a)$ right? $\endgroup$ Commented Apr 25, 2018 at 20:37
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    $\begingroup$ Yes, that's right. :) $\endgroup$ Commented Apr 25, 2018 at 21:01
  • $\begingroup$ @TedShifrin So it means we only need to calculate the first fundamental form at the intersection point and that makes $u$ gone? In particular of the problem, if $\alpha (t_0)=\beta(s_0)=\varphi(u_0, v_0)$ then we have $G=u_0^2 +1$. Am I correct sir ? $\endgroup$
    – RopuToran
    Commented Jun 15, 2021 at 15:11
  • $\begingroup$ @RopuToran I don't know what your "and that makes $u$ gone" means. But, yes, you need $E,F,G$ at the intersection point(s). $\endgroup$ Commented Jun 15, 2021 at 15:33
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    $\begingroup$ @TedShifrin Sorry for my confusing expression. What I mean was when we substitute $u$ with value $u_0$ on $G$, the cosine function becomes a number, not a function anymore. Thank you! $\endgroup$
    – RopuToran
    Commented Jun 16, 2021 at 4:56

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