Use the law of cosines to solve the problem. You must solve for BC first. Solve this problem in order.
A ship travels due west for 94 miles. It then travels in a northwest direction for 119 miles and ends up 173 miles from its original position. To the nearest tenth of a degree, how many degrees north of west (x) did it turn when it changed direction? Show your work.
To solve this problem, we can use the law of cosines to find the length of the side BC first, then use the law of sines to calculate the angle x.
Let's label the angles of the triangle as A, B, and C, and the sides opposite them as a, b, and c, respectively. In this case, we know that angle B is the angle we're trying to find (x degrees), side a is 94 miles, side b is 119 miles, and side c is 173 miles.
According to the law of cosines, we have the following formula:
c^2 = a^2 + b^2 - 2ab * cos(C)
We know that side a is 94 miles, side b is 119 miles, and side c is 173 miles. Substituting these values into the formula, we get:
173^2 = 94^2 + 119^2 - 2 * 94 * 119 * cos(C)
Using a calculator to solve for cos(C), we find:
cos(C) = (94^2 + 119^2 - 173^2) / (2 * 94 * 119)
cos(C) ≈ -0.873
To find angle C, we can take the inverse cosine (cos^(-1)) of -0.873:
C = cos^(-1)(-0.873)
C ≈ 150.4 degrees
Since angle C is the angle between sides a and c, which represents the direction of motion when the ship changed direction, we have C = x degrees north of west.
Therefore, x ≈ 150.4 degrees north of west.
If the ship started at A, turned at B, and ended up at C, then we want to find angle B.
173^2 = 94^2 + 119^2 - 2*94*119 cosB
cosB = -1733/5593
B = 108°
so the ship turned x=72° north at B