The angle is\(\theta = 10^\circ \).

The number of sides of the dome is\(N = 36\).

The required force on each segment is \(F = 4.2 \times {10^5}\;{\rm{N}}\).

In this problem, while evaluating the tension that must exist in each segment, write Newton’s second law for the horizontal as well as vertical directions.

The following is the free-body diagram.

The relation to calculate the force can be written as:

\(\begin{array}{c}\sum {F_{\rm{x}}} = 0\\\left( {{F_{\rm{2}}} \times \cos \theta } \right) - \left( {{F_{\rm{1}}} \times \cos \theta } \right) = 0\\{F_{\rm{1}}} = {F_{\rm{2}}}\end{array}\)

Here, \({F_{\rm{1}}}\)and \({F_{\rm{2}}}\) are the tension forces on each segment of the dome.

The relation to calculate the tension force can be written as:

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\\left( {{F_{\rm{1}}} \times \sin \theta } \right) + \left( {{F_{\rm{2}}} \times \sin \theta } \right) - {F_{\rm{B}}} = 0\end{array}\)

On plugging the values in the above relation, you get:

\(\begin{array}{c}\left( {{F_{\rm{1}}} \times \sin \theta } \right) + \left( {{F_{\rm{1}}} \times \sin \frac{\theta }{2}} \right) - {F_{\rm{B}}} = 0\\{F_1} = \frac{{{F_{\rm{B}}}}}{{2\sin \frac{\theta }{2}}}\\{F_1} = \left( {\frac{{4.2 \times {{10}^5}\;{\rm{N}}}}{{2\sin \frac{{10^\circ }}{2}}}} \right)\\{F_1} = 2.409 \times {10^6}\;{\rm{N}}\end{array}\)

Thus, \({F_1} = 2.409 \times {10^6}\;{\rm{N}}\) is the required force.