Let a and b are two vectors inclined at an angle of 60°. If $|\vec a|=|\vec b|=2$, then the angle between $\vec a$ and $\vec a + \vec b$, is

30°

We have,

$|\vec a|=|\vec b|=2$ and $\vec a.\vec b=|\vec a||\vec b|\cos 60°=2$

Now,

$|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2(\vec a.\vec b)$

$⇒|\vec a+\vec b|^2=4+4+4$

$⇒|\vec a+\vec b|=2\sqrt{3}$

Let θ be the angle between a and a + b. Then,

$\cos θ=\frac{\vec a.(\vec a+\vec b)}{|\vec a||\vec a+\vec b|}=\frac{\vec a.\vec a+\vec a.\vec b}{|\vec a||\vec a+\vec b|}=\frac{4+2}{2×2\sqrt{3}}=\frac{\sqrt{3}}{2}$

$⇒θ=30°$