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Determine the manitude and coordinate direction angles of $\mathrm{F}_{2}$ so that the resultant of the two forces is zero.

$\stackrel{+x}{\longleftarrow}$

$+z \uparrow$

$xy{\longrightarrow}$

$\stackrel{x}{\longleftarrow} \sum F_{x}=0.$

$180 \cos 15 \sin 60+F_{2 x}=0$

$∴ F_{2 x}=150.6 N$

$\stackrel{+}{\longrightarrow} \sum F_{y}=0.$

$180 \cos 15 \cos 60+F_{2 y}=0$

$∴ F_{2y} =-86.9 N$

$+\uparrow \ \sum F_{2}=0$

$-180 \sin 15+F_{2 z}=0$

$∴ F_{2z} =46.6 N$

$\overrightarrow {F_{2}}=(-150.6 i-86.9 j+46.6 k) N$

$\left|F_{2}\right|=180 \mathrm{N}$

$\begin{array}{l}{\alpha=\cos ^{-1}\left(\frac{-150.6}{180}\right)=147^{\circ}} \\ {\beta=\cos ^{-1}\left(\frac{-86.9}{180}\right)=119^{\circ}} \\ {\gamma=\cos ^{-1}\left(\frac{46.6}{180}\right)=75^{\circ}}\end{array}$