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$(A)$ To find the resultant force, we resolve each force into its $X$ and $Y$ components.$X$-components:$\Sigma F_x = 19 + 15 \cos 60^{\circ} - 16 \co

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$6.5 \,N, 330^{\circ}$

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$(A)$ To find the resultant force, we resolve each force into its $X$ and $Y$ components.$X$-components:$\Sigma F_x = 19 + 15 \cos 60^{\circ} - 16 \cos 45^{\circ} - 11 \cos 30^{\circ}$$\Sigma F_x = 19 + 15(0.5) - 16(0.707) - 11(0.866)$$\Sigma F_x = 19 + 7.5 - 11.312 - 9.526 = 5.662 \,N$$Y$-components:$\Sigma F_y = 15 \sin 60^{\circ} + 16 \sin 45^{\circ} - 11 \sin 30^{\circ} - 22$$\Sigma F_y = 15(0.866) + 16(0.707) - 11(0.5) - 22$$\Sigma F_y = 12.99 + 11.312 - 5.5 - 22 = -3.198 \,N$Resultant force $R = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2} = \sqrt{(5.662)^2 + (-3.198)^2} \approx \sqrt{32.06 + 10.23} \approx \sqrt{42.29} \approx 6.5 \,N$.Direction $	heta = 	an^{-1}\left(\frac{\Sigma F_y}{\Sigma F_x}ight) = 	an^{-1}\left(\frac{-3.198}{5.662}ight) \approx 	an^{-1}(-0.565) \approx -29.5^{\circ} \approx 330.5^{\circ}$.Thus, the resultant is approximately $6.5 \,N$ at $330^{\circ}$.

Consider an object kept at the centre, in the $XY$-plane, on which five coplanar forces act as shown in the figure. The resultant force on the object is

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