Given are 2 right angle triangles. $\triangle PAB$ and $\triangle PBC$ such that $PA=16$, $AB=12$ and $BC=21$. Express the value of cos $\angle APC$ in the form $\frac{x}{y}$, where x,y are integers. $\triangle PBC$ sits on $\triangle PAB$ such that $PB$ is the hypotenuse of $\triangle PAB$ and the base of $\triangle PBC$.
I have found $PB$ by pythagoras to be 20 and used this to determine $PC$ to be 29. To express $\angle PAC$ it is possible to combine $\angle APB$ and $\angle BPC$ like this; $cos(\frac{16}{20}) + cos(\frac{20}{29})$ or is the another way?
Draw the triangle. Common vertex is at $P.$
No, $\cos(p1+p2)\ne\cos p1 + \cos p2 $. We have
$$ \cos \angle APC= \cos(p1+p2)=\cos p1 \cos p2-\sin p1\sin p2$$
$$= \dfrac{ 16 }{20 } . \dfrac{ 20 }{ 29 }- \dfrac{12 }{ 20 } . \dfrac{21 }{29 } = \dfrac{ 17 }{145 } .$$