Question 1106105
I agree that to the nearest degree {{{D=41^o}}} .
From there, you could use the trigonometric ratios {{{sin(D)=AE/AD}}} or {{{cos(D)=DE/AD}}} .
You could say that
{{{sin(D)=AE/AD}}}
{{{sin(41^o)=7ft/AD}}}
and using the approximate value  {{{sin(41^o)=0.656}}}
{{{0.656=7ft/AD}}}
{{{0.656AD=7ft}}}
{{{AD=7ft/0.656}}}
{{{AD=10.67ft}}}(rounded)
To the nearest unit {{{AD=highlight(11ft)}}} .
 
The problem makes sense if you are only able to use trigonometric value tables,
just like I did before Hewlett Packard started selling calculators.
With a calculator, you could give your answers with more decimal places,
and you may even be tempted to calculate AD using the Pythagorean theorem, as
{{{AD=sqrt(AE^2+DE^2)=sqrt((7ft)^2+(8ft)^2)=sqrt(49ft^2+64ft^2)=sqrt(113ft^2)=10.63ft}}}