Two dogs pull horizontally on ropes attached to a post; the angle problem 1 between the ropes is 60.0 degrees. If dog A exerts a force of 270N and dog B exerts a force of 300N find the magnitude of the resultant force and the angle it makes with dog A's rope.

You have a SAS triangle problem. You need the third side, and an angle.

Draw this out in a head to tail vector on vector.

one vector 270 at zero degrees (dogA). Next, starting at the end of that vector, a 300N at 60 deg upward (the angle then in the triangle is 120 deg).

First, the length of the other side:
Law of cosines..
R^2=a^2+b^2 -2abcos120
you know a, b find R

Then the angle between the dogA and the resultant.
SinAngle/opposite side=Sin120/R
where the opposite side in this triangle is 300
solve for the angle.

F = 270[0o] + 300[60o]
F = 270+300*Cos60 + 300*sin60=420+259.8i
= 494N.[31.74o].

F = 31.74o above dog A.

F=
270
2
+300
2
+2(270)(300)cos60
​
=494 N
\theta=\arctan{\frac{(300)\sin{60}}{270+(300)\cos{60}}}=31.7\degreeθ=arctan
270+(300)cos60
(300)sin60
​
=31.7°

F=
270
2
+300
2
+2(270)(300)cos60
​
=494 N
\theta=\arctan{\frac{(300)\sin{60}}{270+(300)\cos{60}}}=31.7\degreeθ=arctan
270+(300)cos60
(300)sin60
​
=31.7°