Consider the helicoid $S$ given by the parametrization
$$x(u,v)=(v\cos u,v\sin u,u).$$
a) Let $T$ be the curvilinear triangle on $S$ which is the image under $x$ of the triangle
$\{(u,v): 0 \leq u \leq a, 0 \leq v \leq \sinh u\}$ for some $a > 0$.
Find the angles of $T$, the length of its edges, and its area.
b) Find the asymptotic curves on $S$.
c) Find the Gaussian and the mean curvature of $S$.
I have a general idea of how to solve b) and c), but I am struggling with a). I think arc-length formalization should give the length but I'm not sure about the angles of $T$ and the area.
Angles: Consider the curves $$ \begin{align*} \gamma_1&\colon [0,a]\to \mathbb{R}^2: t\mapsto (t,0) \\ \gamma_2&\colon [0,\sinh(a)]\to\mathbb{R}^2 :t \mapsto (a,t) \\ \gamma_3&\colon [0,a]\to\mathbb{R}^2 : t \mapsto (a-t,\sinh(a-t)). \end{align*} $$ Then the three curves $c_i(t)=x(\gamma_i(t))$, $i=1,2,3$ form the triangle on the surface. As an example, I'll calculate one angle. Note that $c_1'(a)=(0,0,1)$ and $c_2'(0)=(\cos a,\sin a, 0)$. Using the formula $$ \cos \theta = \frac{c_1'(a)\cdot c_2'(0)}{\|c_1'(a)\| \|c_2'(0)\|} $$ one finds that the exterior angle between the two curves is $\frac{\pi}{2}$. If one wants to find the interior angle, one has to substract this angle from $\pi$. But for this specific angle this is $\pi-\frac{\pi}{2}=\frac{\pi}{2}$ as well.
Lengths: You are right. In order to calculate the lengths one uses the formula for the arclength: $$ \int_0^{\text{upper bound}} \left\|c_i'(t)\right\|\,dt. $$
Area: For the area one calculates the surface integral of the constant function $1$ on the triangle. Filling in the parametrisation and boundaries gives $$ \int_{0}^a \int_0^{\sinh u} \|x_u(u,v)\times x_v(u,v)\|\,dv du =\int_{0}^a \int_0^{\sinh u} \sqrt{1+v^2}\,dv du. $$ I am not going to spoil the answer for you. I only want to mention that the integral is pretty nice, because the substitution $v=\sinh t$ will do the trick.