Find the height of a trapezoid 
A trapezoid $ABCD$ is given $(AB$ $||$ $CD)$ with side lengths $AB=18,BC=\sqrt{74},CD=5$ and $AD=\sqrt{61}.$ Find the sines of $\measuredangle A$ and $\measuredangle B.$


(I have only studied trig functions of acute angles)
Since we are given the four sides of the trapezoid, it is enough to find the height. I am not sure how to approach the problem. If $CP$ is parallel to $AD,$ we can find the area of triangle $PBC$ by Heron's formula (the area of a triangle when the length of all three sides are known) but we still haven't studied it. Can you give me a hint for another solution? Thank you in advance!

First Solution
$PC=AD=\sqrt{61}$, $BC=\sqrt{74}$, $BP=13$
$cos \angle{B}=\frac{74+169-61}{26\sqrt{74}}=\frac{7}{\sqrt{74}}$
$sin \angle{B}=\frac{5}{\sqrt{74}}$
$cos \angle{A}=  cos \angle{P}=\frac{61+169-74}{26\sqrt{61}}=\frac{6}{\sqrt{61}}$
$sin\angle{A}=\frac{5}{\sqrt{61}}$
Second Solution
Let $AD_1=x, BC_1=13-x, CC_1=DD_1=h$
$h^2=61-x^2=74-(13-x)^2$
$x=6, h=5$
Third Solution
$BP=13$ since $AP=CD=5$
$CP=AD=\sqrt{61}$
Let $PC_1=x$,  $BC_1=13-x$,  $CC_1=h$
$h^2=61-x^2=74-(13-x)^2$
$x=6, h=5$
