Let's label the sides of the triangle:
AB = 94 miles (due west)
AC = 119 miles (northwest)
BC = 173 miles (resultant displacement)
Now, let's use the Law of Cosines to solve for BC:
BC^2 = AB^2 + AC^2 - 2(AB)(AC)cos(x)
BC^2 = 94^2 + 119^2 - 2(94)(119)cos(x)
BC^2 = 8836 + 14161 - 22372cos(x)
BC^2 = 22997 - 22372cos(x)
Given that BC = 173 miles, we can now solve for cos(x):
173^2 = 22997 - 22372cos(x)
29929 = 22997 - 22372cos(x)
-6974 = -22372cos(x)
cos(x) = 0.3113
To find x, we take the inverse cosine of 0.3113:
x = cos^(-1)(0.3113)
x ≈ 71.6 degrees
Therefore, the ship turned approximately 71.6 degrees north of west.
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?
Use the Law of Cosines to solve the problem. You must solve for BC first. Solve this problem in order.
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1 answer