The problem is modelled as right angle triangle, shown in the diagram below
The distance between the pilot and the house is labelled as [tex]x[/tex]
Using the trigonometry ratio of sine
[tex]sin(31)= \frac{opposite}{hypotenuse} [/tex]
[tex]sin(31)= \frac{2010}{x} [/tex]
[tex]x= \frac{2101}{sin(31)} = 4079.3 ft[/tex]
