An explorer walks 22 km in a northerly direction and then walks in a direction 60 degrees south of east for 47 km. How far is she from where she started?

Suppose explorer start from origin $A$
$A$ to north $22$ $Km$ to a point $B$
$B$. Then Move $60^\circ$ South East $47$ $Km$ to reach a point $C$
$C$. Join point $AC$,
$AC$, Length of $AC$
$AC$ would be displacement from origin
Now draw a line $DC$ from point $C$ that meet line $AB$ on point $D$
$D \triangle ABC$
$\triangle ABC$ Given $\angle B = 60^\circ$
$\angle B = 60^\circ$ $AB = 22$ $Km$
$AB = 22$ $Km$ $BC = 47$ $Km$
$BC = 47$ $Km$
Displacement $= AC = ?$
From $\triangle DBC$
$\triangle DBC \hspace{2mm} Sin\hspace{2mm} B = \frac{DC}{BC}$
$Sin\hspace{2mm} 60^\circ = \frac{DC}{47}$
$DC = 47 \times Sin\hspace{2mm} 60^\circ$
$DC \approx 40.66$ $Km$
From $\triangle DBC$
$\triangle DBC \hspace{2mm} Tan\hspace{2mm} B = \frac{DC}{BD}$
$Tan\hspace{2mm} 60^\circ = \frac{DC}{BD}$
$BD = \frac{DC}{Tan\hspace{2mm} 60^\circ}$
$BD \approx 23.43$ $Km$
$AD = BD - AB$
$AD \approx 23.43 - 22$
$AD \approx 1.43$ $Km$
From $\triangle ADC$
$AC^2 = DC^2 + AD^2$
$AC^2 = 40.66^2 + 1.43^2$
$AC \approx \sqrt{40.66^2 + 1.43^2}$
$AC \approx 40.68$ $Km$
Displacement $AC \approx 40.68$ $Km$.