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

a man looks up to the top of a building. he is 3 meters from the base of the building and the building is 20 meters tall. find the mans distance from his point to the top of the building 'x'.

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