Answer:
distance of 2nd team from 1st team will be: 58.2
Direction of 2nd team from 1st team will be: 14.90 deg North of east
Explanation:
ASSUME Vector is R and makes angle A with +x-axis,
therefore component of vector R is


From above relation
Assuming base camp as the origin, location of 1st team is
away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)


location of 2nd team is at
, at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)


Now position of 2nd team with respect to 1st team will be given by:


Using above values:


distance of 2nd team from 1st team will be:


Direction of 2nd team from 1st team will be:
![Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]](https://tex.z-dn.net/?f=Direction%20%3D%20tan%5E%7B-1%7D%20%5Cfrac%7BR_%7B3y%7D%7D%7BR_%7B3x%7D%7D%20%3D%20tan%5E%7B-1%7D%5B%20%5Cfrac%7B13.71%7D%7B51.49%7D%5D)
Direction = 14.90 deg North of east