TIE - Applying Law of Sines and Cosines Lesson

Applying Law of Sines and Cosines

We can use Law of Sines and Cosines to solve real-world trigonometric applications. Let's try one together:

To determine the distance between two points A and B, a surveyor choose a point C that is 383 yards from A and 548 yards from B. If equation image indicator $LaTeX: \angle BAC$ $\angle BAC$ has a measure of 49.5 °, approximate the distance between A and B.

Step One: Draw a picture to represent the situation.

triangle ABC with sides of 383 yd, 548 yd, and 49.5°

Step Two: Solve the triangle. Notice that you have two sides and a non-included angle - so we must use Law of Sines, but check for the ambiguous case!

equation image indicator $LaTeX: \frac{548}{\sin\left(49.5\right)}=\frac{383}{\sin B}\\ 383\sin\left(49.5\right)=548\sin B\\ 0.53145161=\sin B\\ B=32.1°\:or\:B=180-32.1=147.9°$ $\frac{548}{\sin\left(49.5\right)}=\frac{383}{\sin B}\\ 383\sin\left(49.5\right)=548\sin B\\ 0.53145161=\sin B\\ B=32.1°\:or\:B=180-32.1=147.9°$

equation image indicator $LaTeX: \angle B$ $\angle B$ couldn't be LaTeX: 147.9° $147.9°$ because that is too large for the triangle, so now we know it is LaTeX: 32.1° $32.1°$ .

Once we know equation image indicator $LaTeX: \angle B$ $\angle B$ , we know $LaTeX: \angle C$ $\angle C$ must be LaTeX: 180-32.1-49.5=98.4° $180-32.1-49.5=98.4°$

triangle ABC with sides of 383 yd, 548 yd, and angle B 49.5°, angle C 98.4°, and angle A 49.5°

Now that we know all of the angles we can use the Law of Sines to solve for the last side:

equation image indicator $LaTeX: \frac{c}{\sin\left(98.4\right)}=\frac{548}{\sin\left(49.5\right)}\\ 548\sin\left(98.4\right)=c\sin\left(49.5\right)\\ 712.95=c$ $\frac{c}{\sin\left(98.4\right)}=\frac{548}{\sin\left(49.5\right)}\\ 548\sin\left(98.4\right)=c\sin\left(49.5\right)\\ 712.95=c$

So, we can estimate that the distance between A and B is about 713 yards.

triangle ABC with sides of 383 yd, 548 yd, and angle B 49.5°, angle C 98.4°, and angle A 49.5° with side AB of 712.94 yd

Watch this video to try another problem.

One of the most common uses for Trigonometry is in navigation. We've mainly discussed angles in standard position (angles are measured beginning at the positive x-axis and rotate counter-clockwise). However, in navigation, angles are measured using headings. A bearing is also measured from due north.

If I say a boat is heading in the direction 150°, then you measure from due north, 150° clockwise. See the image below:

compass circle with angle denoting 150° rotation

Try these problems to see if you understand:

Watch this video to see how to apply navigational direction and Law of Cosines.

Let's try another problem.

You flew for 22 miles in the direction of 225°, then turned to a direction of 170° for 30 miles. How far are you from your starting point? What direction must you fly to get back to where you started?

1. Draw a picture of what you know.

triangle with sides labelled x, 30 mi, and 22 mi and angles 170° and 45°

2. Now, we have enough information to determine the angle across from side x. By using parallel line relationships, we can determine that angle to be equation image indicator LaTeX: 170°-45°=125° $170°-45°=125°$ .

3. So now, we have two sides and the included angle - so let's use Law of Cosines to solve for the unknown side:

equation image indicator $LaTeX: x^2=\left(22\right)^2+\left(30\right)^2-2\left(22\right)\left(30\right)\cos125\\ x=\sqrt[]{\left(22\right)^2+\left(30\right)^2-2\left(22\right)\left(30\right)\cos125}\\ x\approx46.27$ $x^2=\left(22\right)^2+\left(30\right)^2-2\left(22\right)\left(30\right)\cos125\\ x=\sqrt[]{\left(22\right)^2+\left(30\right)^2-2\left(22\right)\left(30\right)\cos125}\\ x\approx46.27$

4. So, you are about 46.27 miles from your starting point.

5. Now, we need to determine the angle using Law of Sines. We are looking for the angle down at the bottom of the figure.

triangle with sides labelled x, 30 mi, and 22 mi and angles 170° and 45° with 10 and 45° denoted

equation image indicator $LaTeX: \frac{22}{\sin A}=\frac{46.27}{\sin125}\\ sin A=\frac{22\sin125}{46.27}\\ sin A=0.3895\\ A\approx22.92°$ $\frac{22}{\sin A}=\frac{46.27}{\sin125}\\ sin A=\frac{22\sin125}{46.27}\\ sin A=0.3895\\ A\approx22.92°$

6. Using parallel line relationships, we can determine the direction of the travel path is 12.92°.

triangle with sides labelled x, 30 mi, and 22 mi and angles 170° and 45° with 10 and 45° denoted and 22.92° and 12.92° denoted on the last angle

Here are a few practice problems:

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