Can I get some help with problem 2.53 shown below?
let $\displaystyle \theta$ be the angle AB makes with the horizontal
$\displaystyle \phi$ be the angle BC makes with the horizontal
$\displaystyle T$ = wire tension
$\displaystyle W$ = weight of aerailist
$\displaystyle \sum{F_x} = 0$
$\displaystyle (T+35)\cos{\theta} = T\cos{\phi}$
$\displaystyle \sum{F_y} = 0$
$\displaystyle (T+35)\sin{\theta} + T\sin{\phi} = W$
let the length of $\displaystyle AB = x$ ... then the length of $\displaystyle BC = 8-x$
let $\displaystyle h$ = vertical distance (amount of "dip") point B is below AC
$\displaystyle h^2 = x^2 - 2.5^2$
$\displaystyle h^2 = (8-x)^2 - 5.45^2$
$\displaystyle (8-x)^2 - 5.45^2 = x^2 - 2.5^2$
solve for $\displaystyle x$, then you can determine
$\displaystyle \cos{\theta} = \frac{2.5}{x}$
$\displaystyle \cos{\phi} = \frac{5.45}{8-x}$
finally, solve your system of force equations for T and W.