Question 973076
If want cosine, h for hypotenuse,
{{{22^2+14^2=h^2}}}
{{{h=sqrt(22^2+14^2)}}}


{{{cos(theta)=adjascent/hypotenuse}}}


{{{cos(theta)=22/sqrt(22^2+14^2)}}}
Simplify or compute to whatever accuracy you want and then find arc whose cosine is that computed value for cosine.



{{{cos(theta)=0.84366}}}, or if you want to four places,
{{{highlight(cos(theta)=0.8437)}}}