We must use the trigonometric function.
[tex](a)\\\sin(\angle BAC)=\dfrac{ground}{plank}\\\\\sin49^o=\dfrac{3}{|AC|}\\\\\sin49^o\approx0.7547\\\\\dfrac{3}{|AC|}=0.7547\\\\0.7547|AC|=3\ \ \ |:0.7547\\\\|AC|\approx3.9750\approx4\ ft.[/tex]
We can use the Pythagorean theorem.
[tex](b)\\|BC|^2+3^2=4^2\\\\|BC|^2+9=16\ \ \ |-9\\\\|BC|^2=7\to|BC|=\sqrt7\\\\|BC|\approx2.6458\\\\|BC|\approx3\ ft.[/tex]
