Question

Three vectors A, B and C are such that

Answer 1

Since the magnitudes satisfy the identity for right triangles

$A^2=B^2+C^2$

we deduce that B and C are perpendicular, and that ABC is a right triangle (see picture).

So, we can get the angle between A and C using the sine theorem:

$\dfrac{A}{\sin(\hat{A})}=\dfrac{B}{\sin(\hat{B})}$

And we have

$A=5,\quad B=4,\quad \hat{A}=90$

So, we have

$\sin(\hat{B})=\dfrac{B\sin(\hat{A})}{A}=\dfrac{4\sin(90)}{5}=\dfrac{4}{5}$

Which implies

$\hat{B}=\arcsin\left(\dfrac{4}{5}\right)\approx 53.13$