Answer:
18.7 feet far
Step-by-step explanation:
Based on the information given we can conclude that our set up will be a Right triangle (i.e. one angle is 90°), where the hypotenuse will be denoted by the guy wire of
and angle of 21° is the angle formed between then hypotenuse and the base (i.e. ground).
Thus we want to find the base length (lets call it
) of this triangle, so we can use trigonometry as we have one angle and the hypotenuse, as follow:

So the base is approximately 18.7 feet far from the tree of the anchored wire.
Length of base of the top triangle = 23 inches
Length of height of the top triangle = 47 inches
Area of the top triangle :





Thus, area of the top triangle = 540.5 sq.inches
Length of base of the bottom triangle = 47 inches
Length of height of the bottom triangle = 46 inches
Area of the bottom triangle :




Area of the bottom triangle = 1081 sq.inches
Area of the total figure :
= Area of top triangle + Area of bottom triangle


Therefore, the area of the figure = 1621.5 sq.inches
Yes he has enough inches of wire