Answer 1

find the vector components for orthogonal axes x and y, then add x resp y components

for x)
440m . cos 50° + 580m . cos 185° = 282,83 - 577,79 = -294,96 m
for y)
440m . sin 50° + 580m . sin 185° = 286,51m
now Pythagoras rule A² = B² + C²
\[A = \sqrt{\left( B² + C² \right)} = 411,21m\]
and inverted tangent
\[\theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{286,51m}{-294,96m} = -44,2 degrees + n.180°\]
As the direction is in second quadrant, i add 180° to the angle:
-44,2°+180° = 135,8°