Answer
$(a)\space 9.4\space ft$
$(b)\space 69^{\circ}$
Work Step by Step
The performer walks out on the wire a distance of d, and the vertical distance to the net is h. So we can get the magnitude of the displacement by using the Pythagorean theorem as follows,
$S=\sqrt {d^{2}+h^{2}}$
$d=\sqrt {S^{2}-h^{2}}=\sqrt {(26.7\space ft)^{2}-(25\space ft)^{2}}=9.4\space ft$
We can get the angle below horizontal $\theta$ of the displacement by using trigonometry.
$tan\theta=\frac{h}{d}=\gt\theta=tan^{-1}(\frac{h}{d})=tan^{-1}(\frac{25\space ft}{9.4\space ft})=69^{\circ}$
