91Ó°ÊÓ

The tension in the cord is$280\text{ N}$.

The angle made by the rope to the vertical,$\mathrm{θ}=15°$.

The maximum horizontal force, ${\text{F}}_{\text{max}}=93\text{ N}$.

Using the equation for the equilibrium of force, you can find out the weight of the child, the horizontal force the father applied to the child, and the maximum angle made by the rope. The equations are given below.

$\begin{array}{l}\mathbf{W}\mathbf{=}\mathbf{Tcos}\mathbf{}\mathbf{\left(}\mathbf{15}\mathbf{\text{°}}\mathbf{\right)}\mathbf{}\\ \mathbf{F}\mathbf{=}\mathbf{Tsin}\mathbf{}\mathbf{\left(}\mathbf{15}\mathbf{\text{°}}\mathbf{\right)}\mathbf{}\end{array}$

The force equation for the child can be written as,

$\mathrm{W}=\mathrm{Tcos}(15°)$

And

$\mathrm{F}=\mathrm{Tsin}(15°)$

We have, the force equation for weight as,

$\mathrm{W}=\mathrm{Tcos}(15°)$

By using the value of $\text{T}=280\text{ N}$,

we get,

$\text{W}=270.46\text{ N}$

We have, the force equation for the horizontal force

$\mathrm{F}=\mathrm{Tsin}(15°)$

So, we get

$\text{F}=72.47\text{ N}$

We have the force equation for horizontal force as

$\mathrm{F}=\mathrm{°Õ²õ¾\pm ²Ôθ}$

In the vertical direction, we have, the weight of the child,

$\text{W}=270.46\text{ N}$

By rearranging this equation for angle, we get

$\begin{array}{c}\mathrm{θ}=\left(\frac{{\text{F}}_{\text{max}}}{\text{W}}\right)\\ \mathrm{θ}=18.98°\end{array}$