Question

Please Model this real-world scenario with an equation and possibly a sketch.

Answer 1

This can be solved in several ways.  First with the basic trigonometry identity of tangent and sine (or cosine)

tanα=opposite side/adjacent side...

tanα=20/3

α=arctan(20/3)

Then sinα=h/x

x=h/sinα  where α found above and h=20m:

x=20/sin(arctan(20/3)) meters

x≈20.22m  (to nearest one-hundredth of a meter)

...

We could also use the Pythagorean Theorem which is easier for this problem...

x^2=h^2+d^2

x^2=20^2+3^2

x^2=400+9

x=√409 meters

x≈20.22 meters...

...

We could also use the Law of Cosines:

x^2=a^2+b^2-2abcos90  (-2abcos90=0 so we essentially get what we just did above)

x^2=a^2+b^2

x^2=409

x=√409

x≈20.22 meters