Solution

First, let's draw the triangle:

Using the Pythagorean theorem: \[\solve{ h^2 &=& 11^2+13^2 \\ h^2 &=&290\\h&=&\sqrt{ 290 } }\] Here we only take the positive square root since this is a geometric value. To solve for the angles, we will use the inverse trig functions: \[ \solve{\sin^{-1}\left(\frac{{13}}{\sqrt{{290}} }\right) &=& \theta_1\\ \sin^{-1}\left(\frac{{11}}{\sqrt{{290}} }\right) &=& \theta_2\\} \] Make sure your calculator is in degrees and then plug in the values (no need to simplify) and then we have our answers. \[ \solve{ \theta_1 &=& 49.8^\circ\\\theta_2 &=& 40.2^\circ } \]