Answer
$|\overline V|=41$
$\theta=12.68^{\circ}$
Work Step by Step
$\because \overline V=〈a_1,a_2〉=〈40,9〉$
The magnitude of the vector: $|\overline V|=\sqrt{a_1^2+a_2^2}$
$\therefore |\overline V|=\sqrt{40^2+9^2}=\sqrt{1600+81}=\sqrt{1681}=41$
The direction of the vector: $\theta=tan^{-1}\dfrac{a_2}{a_1}$
$\theta=tan^{-1}\dfrac{9}{41}=12.68^{\circ}$
