Write each force in component form:
<em>v </em>₁ : 50 N due east → (50 N) <em>i</em>
<em>v</em> ₂ : 80 N at N 45° E → (80 N) (cos(45°) <em>i</em> + sin(45°) <em>j</em> ) ≈ (56.5 N) (<em>i</em> + <em>j</em> )
The resultant force is the sum of these two vectors:
<em>r</em> = <em>v </em>₁ + <em>v</em> ₂ ≈ (106.5 N) <em>i</em> + (56.5 N) <em>j</em>
Its magnitude is
|| <em>r</em> || = √[(106.5 N)² + (56.5 N)²] ≈ 121 N
and has direction <em>θ</em> such that
tan(<em>θ</em>) = (56.5 N) / (106.5 N) → <em>θ</em> ≈ 28.0°
i.e. a direction of about E 28.0° N. (Just to clear up any confusion, I mean 28.0° north of east, or 28.0° relative to the positive <em>x</em>-axis.)