You can model this situation with a right triangle ABC, where AB is the horizontal leg, and BC is the vertical leg:
[tex]\quad\quad\quad\quad\quad C\\\\A\quad\quad\quad\quad\; B[/tex]
We know that BC=10 and AC=76.
We are interested in angle A.
Using the law of sines, we know that
[tex]\dfrac{BC}{\sin(A)}=\dfrac{AC}{\sin(90)}\iff \dfrac{BC}{\sin(A)}=AC \iff \sin(A)=\dfrac{BC}{AC}=\dfrac{10}{76}[/tex]
So, the angle is
[tex]\sin(A)=\dfrac{10}{76}\implies A=\arcsin\left(\dfrac{10}{76}\right)[/tex]
