Answer:
[tex]F=\frac{32\sqrt{3}}{3}\text{\:or\:} \approx 18.48\:\text{N}[/tex]
Step-by-step explanation:
We can find the horizontal component of force he applied by using the formula for work:
[tex]W=F\Delta x[/tex]
Solving for force, we have:
[tex]160=F\cdot 10,\\F=16\:\text{N}[/tex]
However he applied the original force at an angle of [tex]30^{\circ}[/tex] to the horizontal. This force of 16 newtons is only the horizontal component of that force. To find the magnitude of the original force, we can use basic trigonometry:
[tex]\cos 30^{\circ}=\frac{16}{x},\\x=\frac{16}{\cos 30^{\circ}},\\\\x=\frac{16}{\frac{\sqrt{3}}{2}},\\\\x=16\cdot\frac{2}{\sqrt{3}}=\frac{32}{\sqrt{3}}=\boxed{\frac{32\sqrt{3}}{3}}[/tex]
