Solution 4.6c

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Resultant force \displaystyle =80\textrm{.}9\mathbf{i}-11\textrm{.}3\mathbf{j}\ \text{N}

Thus

\displaystyle \tan \alpha =\frac{-11\textrm{.}3}{80.9}=-0\textrm{.}14

giving

\displaystyle \alpha =-8\textrm{.}0{}^\circ

As the vector is in the fourth quadrant

Image:4.6c.gif