Recall that
[tex]\cos(90-x)^\circ=\sin x^\circ[/tex]
which tells you that [tex]\cos40^\circ=\cos(90-50)^\circ=\sin50^\circ[/tex], so in fact
[tex]\cos40^\circ=\sin50^\circ=\dfrac{10}{15}[/tex]
Attached is an image demonstrating why the identity is true.