Browse Questions

If $1^{\circ}=\alpha$ radians then the approximate value of $\cos(60^{\circ}1')$ is

$(a)\;\frac{1}{2}+\frac{\alpha \sqrt 3}{120} \quad (b)\;\frac{1}{2}-\frac{\alpha}{120} \quad (c)\;\frac{1}{2}-\frac{\alpha \sqrt 3}{120} \quad (d)\;\frac{1}{2}+\frac{\alpha}{120}$