Question

A ship leaves port on a bearing of 28 degrees and travels 8.2 miles. The ship then turns due east and travels 4.3 miles. How far is the ship port? What is the bearing from port?

Answer 1

is that about right?

Answer 2

yes but how do we find the two questions? @jdoe0001

Answer 3

well, notice the line from the "blue" trajectory to the North-South bearing line, it makes a right angle

so you have a 28 degrees angle with a 8.2 side, so to find the "vertical line" side
use the cosine function for the 28 degrees

$$
cos(28^o) = \cfrac{a}{8.2} \implies a = cos(28^o)\times 8.2
$$

Answer 4

now to find the shorter "gray" side, you'd use the sine function
$$
sin(28^o) = \cfrac{b}{8.2} \implies b = sin(28^o)\times 8.2
$$

Answer 5

once you have the "gray" line, you add it to the 4.3 trajectory blue one and you have the whole length for that side in a rigtht triangle

Answer 6

so, the "vertical" black one, that you got from the cosine function or your "a"
the "gray" added to the 4.3, which will be your "b"

and the purple one, will be the hypotenuse
and you can get the purple one, or the distance from the port, using pythagorean theorem

Answer 7

thanks its 10.9 miles :) but how about the second question i dont really understand what its asking

Answer 8

as far as the angle it makes
same, you use
the "vertical" black one, that you got from the cosine function or your "a"
the "gray" added to the 4.3, which will be your "b"


and use those 2 sides, so they're the opposite and adjancent, and use

$$
tan(\theta) = \cfrac{b}{a} \implies tan(\theta) = somevalue\\
tan^{-1}\pmatrix{tan(\theta)} = tan^{-1}(somevalue)
$$

Answer 9

anyhow, need to dash for now :)

Answer 10

thanks so much! its approx. 20 degrees.