Question

Please help! 30 points!

Use the Law of Cosines to solve the problem.

A ship travels due west for 83 miles. It then travels in a northwest direction for 111 miles and ends up 165 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.

Answer 1

Answer: $64.2^{\circ}$

Step-by-step explanation:

According to the cosine rule

$\cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Here, $a=83, b=111, c=165,\ C=180^{\circ}-\theta$

Putting values

$\Rightarrow \cos (180^{\circ}-\theta)=\dfrac{83^2+111^2-165^2}{2\times 83\times 111}\\\\\Rightarrow \cos(180^{\circ}-\theta)=\dfrac{6889+12321-27225}{18426}\\\\\Rightarrow \cos(180^{\circ}-\theta)=\dfrac{-8015}{18426}=-0.4349\\\\\Rightarrow 180^{\circ}-\theta=\cos^{-1}(-0.4349)=115.77^{\circ}\\\Rightarrow \theta=64.23\approx 64.2^{\circ}$

So, the ship turns by $64.2^{\circ}$ north of west

Answer 2

The ship turned an at angle of 64.2 degrees north of west

Represent the angle, the ship turned due north of west with theta.

From the figure, we have the following equation

$\cos(A) = \frac{b^2 + c^2 - a^2}{2bc}$

So, we have:

$\cos(A) = \frac{83^2 + 111^2 - 165^2}{2 * 83 *111}$

Evaluate the expression on the right-hand side

$\cos(A) = -0.4350$

Take the arc cos of both sides

$A = \cos^{-1}(0.4350)$

$A =115.8^o$

The value of theta is then calculated as:

$\theta = 180 -A$

This gives

$\theta = 180 -115.8^o$

$\theta = 64.2^o$

Hence, the ship turned an at angle of 64.2 degrees north of west

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