A hiker travels N 33 degrees W for 15km, then turns 90 degrees and walks 12km in the direction S 57 degrees W At that time how far is the hiker from his starting point? What is his bearing from his starting point?


My teacher went over this so fast and I am utterly confused

For this, you can do two things:
One, the pythagorean theorum:
a^2 + b^2 = c^2 where c is your hypothenuse.
However, based off your title, I'd assume you are instructed to solve using the trig functions.
Sin = Opposite/Hypothenuse
Cos = Adjacent/Hypothenuse
Tan = Opposite/Adjacent

It's easy to remember with the acronym:
SohCahToa

57__12___... In this case, you will solve for the diagonal line. X
... \... 90|... You can use Sin33 = 12/X
... \... |... You can use Cos57 = 15/X
... X\... |15... Plug one of these trig functions into your calculator to get a decimal value, then
... \.. |... you simply convert the equation to get X by itself. If you don't have a calculator
... \|33... with trig functions, you may find them in the back of your book.