Answer
$54.13\space km, 11.7^{\circ}$
Work Step by Step
Please see the attached image first.
Here we use the component method to find the straight line distance between the teams.
Let's take,
The vector from the base camp to the first team = A
The vector from the base camp to the second team = B
The position vector from the first team to the second team = C
From the image, we can get,
$B=A+C=\gt C=B-A$
So,
$C_{x}=B_{x}-A_{x}$ ; Let's plug known values into this equation.
$C_{x}=29\space km\times sin35^{\circ}-(-38\space km)\times cos 19^{\circ}=53\space km$
$C_{y}=B_{y}-A_{y}$ ; Let's plug known values into this equation.
$C_{y}=29\space km\times cos35^{\circ}-38\space km\times sin 19^{\circ}=11\space km$
By using the Pythagorean theorem, we can get.
$C=\sqrt {C_{x}^{2}+C_{y}^{2}}=\sqrt {(53\space km)^{2}+(11\space km)^{2}}=54.13\space km$
By using trigonometry , we can get.
$tan\theta=\frac{C_{y}}{C_{x}}=\frac{11\space km}{53\space km}$
$\theta=tan^{-1}(0.21)=11.7^{\circ}$
