Let x be the angle of the smaller right triangle and y be the angle of the larger right triangle so that:

\(\displaystyleϕ={y}−{x}\)

For the smaller right triangle with hypotenuse as guy wire B, we can use the tangent ratio to find x:

\(\displaystyle{\tan{{x}}}=\frac{{10}}{{40}}\)

\(\displaystyle{x}={{\tan}^{{-{{1}}}}{\left(\frac{{10}}{{40}}\right)}}\)

\(\displaystyle{x}\sim{14.04}°\)

For the larger right triangle with hypotenuse as guy wire A, we can use the tangent ratio to find y:

\(\displaystyle{\tan{{y}}}=\frac{{{30}+{10}}}{{40}}\)

\(\displaystyle{\tan{{y}}}={1}\)

\(y=\tan^{-1}*1\)

\(y=45^{\circ}\)

Hence, \(\displaystyleϕ={45}°−{14.04}°\)

\(\displaystyleϕ≈{30.96}°\)