We know, that for any point M on the terminal side of angle:
cosθ = (x coordinate of M) / (distance from M to the origin)
So:
[tex]x=16\\\\d=\sqrt{16^2+16^2}=\sqrt{2\cdot16^2}=\sqrt{2}\cdot\sqrt{16^2}=16\sqrt{2}\\\\\\\cos\theta=\dfrac{x}{d}=\dfrac{16}{16\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\boxed{\dfrac{\sqrt{2}}{2}}[/tex]
Answer B.
