Question

Evaluate cos(sin^-1(5/13)). I know the graph of the triangle is in Quadrant I and arcsin(5/13) is sin(x)=5/13. How do I solve from there?

Answer 1

sin(x)=5/13 <--- that's correct

so that means \(\bf sin(\theta)=\cfrac{5}{13}\to \cfrac{opposite}{hypotenuse}\to \cfrac{b}{c}\)

Answer 2

\(\bf c^2={\color{blue}{ a}}^2+b^2\implies \sqrt{c^2-b^2}={\color{blue}{ a}}\)