A Ship travels due west for 64 miles. It then travels in a northwest direction for 104 miles and ends up 136 miles from its original position. To the nearest tenth of a degree, how many degrees north of west did it turn when it changed direction? Show your work.
To understand the focus of the question a diagram would help:
The ship travels towards the West (left) and then north west (ie it turns diagonally to the right) creating an angle. The angle we are looking for is the acute angle created if the ship had kept going west (left) and then looked back towards the north.
We have the lengths of the three lines and can create a triangle with base 64 miles, then side (to the north west) 104 miles and opposite the angle (back from the end point north west to the starting point) 136 miles.
We can use the cos rule of `a^2=b^2 +c^2 -2bc cos A`
A is the angle we will calculate from which we can find the acute angle from west back to north west (ie the degrees north OF west) by: `180 - A`
`therefore 136^2= 64^2+104^2- 2 (64)(104)cosA`
(remember that the side a will be opposite the angle A)
`18496=4096+10816 - 13312cos A`
`therefore (18496-4096-10816)/-13312 =cos A`
`therefore cos A=105.6` degrees
and `180-105.6=74.4` degrees
Ans: therefore the ship turned 74.4 degrees north of west when it changed direction.