Answer
$7.1\space m, 9.9^{\circ}\space north\space to\space east$
Work Step by Step
Please see the attached image first.
The resultant $R$ is the displacement that would have been needed to "hole the ball" on the very first putt. We can find $R$ by calculating the x, and y components of it. Let's take the +x & +y axis along the directions of east & north respectively.
$\rightarrow R_{x}=A+Bcos20^{\circ}$ ; Let's plug known values into this equation.
$R_{x}=5\space m+(2.1\space m)cos20^{\circ}=7\space m$
$\uparrow R_{y}=Bsin20^{\circ}+C$ ; Let's plug known values into this equation.
$R_{y}=(2.1\space m)sin20^{\circ}+0.5\space m=1.22\space m$
The magnitude of $R=\sqrt {R_{x}^{2}+R_{y}^{2}}=\sqrt {(7\space m)^{2}+(1.22\space m)^{2}}=7.1\space m$
From trigonometry,
$tan\theta=\frac{1.22\space m}{7\space m}=\gt\theta=tan^{-1}(\frac{1.22\space m}{7\space m})=9.9^{\circ}$
So, the required direction is $9.9^{\circ}$ North of east
