Theorem 7.3.1. Law of Sines.
For a triangle with angles and sides as labeled in Figure 7.3.2,
\begin{equation*}
\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}.
\end{equation*}
Section 7.3 Non-Right Triangles
This section addresses the following topics.
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Read and use mathematical models in a technical document
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Communicate results in mathematical notation and in language appropriate to the technical field
This section covers the following mathematical concepts.
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Use models including linear, quadratic, exponential/logarithmic, and trigonometric (skill)
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Analyze non-right triangles (skill)
In Section 7.1 we learned about relationships between angles of the triangles and their sides. However, most of our work was restricted to right triangles. This section demonstrates how to make similar calculations on non-right triangles.
The one relationship that did not require a right angle is the Triangle Angle Sum Theorem.
Subsection 7.3.1 Law of Sines
In Section 7.1 we saw that there was a relationship between angles of a triangle and the side ratios. More generally there is a relationship between the magnitude of an angle of a triangle and the length of the side opposite it. The following theorem expresses this relationship.
The angles are labeled A, B, and C. The sides opposite angle A, B, and C are labeled a, b, and c respectively.
Example 7.3.3.
A triangle has an angle with measure \(50^\circ\) which is opposite a side of length 6. The triangle has another angle with measure \(45^\circ\text{.}\)
Rounding in this example is arbitrarily chosen to be two decimal places.
(a)
What is the length of the side opposite the \(45^\circ\) angle?
Because we know two angles and a side opposite one of them we can use the law of sines.
\begin{align*}
\frac{\sin(50^\circ)}{6} \amp = \frac{\sin(45^\circ)}{b}. \amp \amp \text{Set up Law of Sines}\\
b \sin(50^\circ) \amp = 6 \sin(45^\circ). \amp \amp \text{Clear the denominators}\\
b \amp = 6 \frac{\sin(45^\circ)}{\sin(50^\circ)}. \amp \amp \text{Divide to solve for }b\\
\amp \approx 5.54.
\end{align*}
(b)
What is the third angle and the length of the side opposite it?
We can use the angle sum fact to calculate the third angle. \(50^\circ+45^\circ+\alpha=180^\circ\) so \(\alpha=85^\circ\text{.}\) As in the previous step we know an angle (50°), the side opposite it (6), and another angle (85°) so we can use the Law of Sines.
\begin{align*}
\frac{\sin(50^\circ)}{6} \amp = \frac{\sin(85^\circ)}{c}. \amp \amp \text{Set up the Law of Sines}\\
c \sin(50^\circ) \amp = 6 \sin(85^\circ). \amp \amp \text{Clear the denominators}\\
c \amp = 6 \frac{\sin(85^\circ)}{\sin(50^\circ)} \amp \amp \text{Divide to solve for }c\\
\amp \approx 7.80.
\end{align*}
If you recall triangle congruency, this was known as Angle-Angle-Side.
Example 7.3.4.
A triangle has two angles with measure 40° and 60.20°. The side between these two angles has length 7.66.
What is the measure of the other angle, and what are the other side lengths? Everything will be rounded to two decimal places.
The third angle measure is the easiest to calculate, because we can use the angle sum theorem. \(40^\circ+60.20^\circ+\alpha=180^\circ\text{.}\) Thus \(\alpha = 79.80^\circ\text{.}\)
Now we know an angle (\(\alpha\)) and the length of the side opposite it (7.66). This enables us to use the Law of Sines to calculate the other two side lengths.
\begin{align*}
\frac{\sin(40^\circ)}{a} & = \frac{\sin(79.80^\circ)}{7.66}. \amp \amp \text{Set up Law of Sines}\\
\frac{a}{\sin(40^\circ)} & = \frac{7.66}{\sin(79.80^\circ)}. \amp \amp \text{Proportion true both ways}\\
a & = \frac{7.66\sin(40^\circ)}{\sin(79.80^\circ)} \amp \amp \text{Multiply to solve}\\
& \approx 5.00281961\\
& \approx 5.00\text{.}
\end{align*}
We calculate the third side the same way.
\begin{align*}
\frac{\sin(60.20^\circ)}{a} & = \frac{\sin(79.80^\circ)}{7.66}.\\
\frac{a}{\sin(60.20^\circ)} & = \frac{7.66}{\sin(79.80^\circ)}.\\
a & = \frac{7.66\sin(60.20^\circ)}{\sin(79.80^\circ)}\\
& \approx 6.75382345\\
& \approx 6.75\text{.}
\end{align*}
For triangle geometry, this was known as Angle-Side-Angle.
Checkpoint 7.3.5.
The following activity demonstrates the relationship between the magnitude of an angle and the length of the side opposite that angle.
Activity 16.
Use the illustration above to answer the questions below.
(a)
Use the slider to set the angle to about 10°.
What happens to the length of the opposite side as you increase the angle to 170°?
(b)
Recall the Triangle Angle Sum relationship. As the angle at A increases what must be happening to the sum of the measures of the other two angles?
What must be happening to each of the other two angles?
(c)
In this illustration the side connecting A and B is remaining the same length. The angle opposite that side is changing as you noted in the previous step.
Combining the results of the previous two steps, what do you think is true about the following. The angle opposite a larger side is bigger/smaller/unrelated to the angle opposite a shorter side.
Subsection 7.3.2 Ambiguous Triangles
Above we calculated angles and side lengths given partial information about a triangle (two angles and a side). This section presents a case (two sides and an angle) that we cannot resolve without additional information.
Example 7.3.7.
A triangle has an angle of measure \(45^\circ\) and the side opposite it is length 4. Another side has length 5. What are the other angles and side lengths?
We can try to use the Law of Sines.
\begin{align*}
\frac{\sin(45^\circ)}{4} \amp = \frac{\sin(\theta)}{5}.\\
5 \cdot \frac{\sin(45^\circ)}{4} \amp = \sin(\theta).\\
\arcsin\left( \frac{5}{4}\sin(45^\circ) \right) \amp = \theta.\\
62.11^\circ \amp \approx \theta.
\end{align*}
Using the triangle angle sum theorem we learn the other angle measure. \(45^\circ+62.11^\circ+\alpha = 180^\circ\) or \(\alpha \approx 72.89^\circ\text{.}\) We use the Law of Sines again to calculate the length of the final side.
\begin{align*}
\frac{\sin(45^\circ)}{4} \amp = \frac{\sin(72.89^\circ)}{c}.\\
\frac{4}{\sin(45^\circ)} \amp = \frac{c}{\sin(72.89^\circ)}.\\
c \amp = 4 \frac{\sin(72.89^\circ)}{\sin(45^\circ)}.\\
c \amp \approx 5.41.
\end{align*}
This gives us a triangle with angles: 45°, 62.11°, and 72.89°; and with side lengths: 4, 5, and 5.41.
However, \(\sin(117.89^\circ)=\sin(62.11^\circ)\text{,}\) that is, \(\arcsin\left( \frac{5}{4}\sin(45^\circ) \right)\) had multiple possible angles. We will repeat the calulcations above using \(117.89^\circ\) as the second angle. The third angle is \(45^\circ+117.89^\circ+\alpha = 180^\circ\) or \(\alpha \approx 17.11^\circ\text{.}\)
\begin{align*}
\frac{\sin(45^\circ)}{4} \amp = \frac{\sin(17.11^\circ)}{c}.\\
c \sin(45^\circ) \amp = 4\sin(17.11^\circ).\\
c \amp = 4 \frac{\sin(17.11^\circ)}{\sin(45^\circ)}.\\
c \amp \approx 1.66.
\end{align*}
Notice we have two, distinct triangles that match the initial angle and side information. They can be seen in Figure 7.3.8. This indicates an ambiguity if all we know is this particular information.
If you recall triangle geometry, this was the case Side-Side-Angle which does not prove congruent triangles.
When using the Law of Sines we will need to restrict ourselves to the two cases for which it works: Angle-Angle-Side and Angle-Side-Angle.
Subsection 7.3.3 Law of Cosines
For right triangles we know the Pythagorean theorem is a relationship between the sides of those triangles. For triangles without a right angle that relationship must be slightly modified. The more general statement is below.
Theorem 7.3.9. Law of Cosines.
For any triangle with side lengths \(a,b,c\) and angle \(C\) which is opposite the side with length \(c\)
\begin{equation*}
c^2 = a^2+b^2-2ab\cos(C)\text{.}
\end{equation*}
Example 7.3.10.
A triangle has sides of lengths 4.00, 5.39, and 6.13. What are the angles?
We can use the Law of Cosines.
\begin{align*}
4^2 \amp = 5.39^2+6.13^2-2(5.39)(6.13)\cos(A).\\
16 \amp = 29.0521+37.5769-66.0814\cos(A).\\
-50.629 \amp = -66.0814 \cos(A).\\
\frac{-50.629}{-66.0814} \amp = \cos(A).\\
0.76616113 \amp \approx \cos(A).\\
\arccos(0.76616113) \amp \approx A.\\
39.98959796^\circ \amp \approx A.\\
40.0^\circ \amp \approx A.
\end{align*}
With an angle, we could now use the Law of Sines, but for practice we will use the Law of Cosines again.
\begin{align*}
5.39^2 \amp = 4^2+6.13^2-2(4)(6.13)\cos(B).\\
29.0521 \amp = 16+37.5769-49.04\cos(B).\\
-24.5248 \amp = -49.04\cos(B).\\
0.50009788 \amp \approx \cos(B).\\
\arccos(0.50009788) \amp \approx B.\\
59.99352413^\circ \amp \approx B.\\
60.0^\circ \amp \approx B.
\end{align*}
Knowing two of the angles we can use the triangle angle sum theorem to calculate the third angle measure. \(40.0^\circ+60.0^\circ+C=180^\circ\) so \(C = 80.0^\circ\text{.}\)
For triangle congruency, this was known as Side-Side-Side.
Example 7.3.11.
A triangle has sides with lengths 5 and 7 and the angle between them is 40°. What are the length of the other side and the measures of the other angles?
Because we have two sides (\(a,b\)) and the angle between them (\(C\)) we can use the Law of Cosines.
\begin{align*}
c^2 \amp = 5^2+7^2-2(5)(7)\cos(40^\circ)\\
c^2 \amp \approx 20.37688900.\\
\sqrt{c^2} \amp \approx \sqrt{20.37688900}.\\
c \amp \approx 4.514076761\\
\amp \approx 4.51.
\end{align*}
Now that we know a side and the angle opposite it, we can use the Law of Sines to calculate the remaining two angles.
\begin{align*}
\frac{\sin(40^\circ)}{4.514076761} \amp = \frac{\sin(A)}{5}.\\
0.71198126 \amp = \sin(A).\\
\arcsin(0.71198126) \amp = A.\\
45.39634537^\circ \amp \approx A.\\
45.4^\circ \amp \approx A.
\end{align*}
Finally we can use the triangle angle sum theorem to calculate the final angle. \(40^\circ+45.4^\circ+B = 180^\circ\) so \(B=94.6^\circ\text{.}\)
For triangle congruency this was known as Side-Angle-Side.
Checkpoint 7.3.12.
Subsection 7.3.4 Using Trigonometric Laws in Applications
When we encounter non-right triangles in applications, we will need to check if we have the information needed to use the Law of Sines or the Law of Cosines.
Example 7.3.13.
A sailboat sails 3341 ft at heading 300° then turns to heading 245° and sails 4051 ft. How far is the sailbot from its starting position? Round to units, because fractions of a foot are not meaningful in the motion of a vehicle.
First, it helps to sketch an image.
In our sketch we see a non-right triangle. We know the lengths of two sides, and we want the length of the third side (distance from original location). From the headings we can calculate the angle between the two sides of known length. This provides side-angle-side which allows us to use the Law of Cosines to calculate the desired distance (length of third side).
We are given the headings for two segments but not an angle. The angle representing the amount of the turn is the change in headings. If we were at 300° and ended up at 245° then we turned left \(300^\circ-245^\circ = 55^\circ\text{.}\)
Now we can apply the Law of Cosines.
\begin{align*}
c^2 & = 3341^2+4051^2-2(3341)(4051)\cos(55^\circ)\\
& = 27572882 - 27068782\cos(55^\circ)\\
& \approx 27572882 - 15526015.51\\
& \approx 12046866.49.\\
c & \approx \sqrt{12046866.49}\\
& \approx 3470.859618\\
& \approx 3471.
\end{align*}
Thus the sailboat ended up 3471 ft from its starting point.
Example 7.3.14.
Two observers are 2.3 miles apart on the ground at the same altitude. At the same time they record the angle at which they saw an aircraft pass. The observer at point A recorded an angle of 8.00° from horizontal. The observer at point B recorded an angle of 34.20° from horizontal. How high above ground was the aircraft?
First, it helps to sketch an image.
A line connects observer A to the airplane. The angle of inclination is labeled as 8.00°. To the right a line connects observer B to the airplane. The angle of inclination is labeled 34.20°. The distance between them is labeled 2.3 miles. To the right of B is a point directly below the airplane labeled C. A dashed line connects it to the airplane and is labeled h.
How do we know that the larger angle is closer to the aircraft? If we think about an aircraft flying toward us, we realize that our head tips up (bigger angle) as it becomes closer. Thus the closer observer will have the larger angle.
In our sketch we see multiple triangles including some right triangles. We want to calculate the height of the aircraft above ground which is part of the small, right triangle with vertices B, C, and the airplane. Initially we know only the bottom angles (34.20° and the right angle) and none of the lengths. If we can calculate the length of the side from B to the airplane (on the left), then we could use the angles and that side length with the Law of Sines to calculate the height.
That left side is also part of a non-right triangle with vertices at points A, B, and the aircraft. The side from A to B has length 2.9 mi, and we know the angles on either side. This means we have angle-side-angle information, so we can apply the Law of Sines to calculate the length of the hypotenuse we need. In particular
\begin{equation*}
\frac{\sin(P)}{2.3}=\frac{\sin(8.00^\circ)}{c}
\end{equation*}
To calculate the angle at the plane (labeled \(P\) above) we will need to use the Triangle Angle Sum theorem. However, the angle we are given is the other side of the non-right triangle. The angle we need now is on the other side of that line, so its measure is \(180^\circ-34.20^\circ = 145.80^\circ = B\text{.}\) Now, b \(8.00^\circ+145.80^\circ+P = 180^\circ\) So, the angle at the aircraft is \(P = 26.20^\circ\text{.}\)
Now we can apply the Law of Sines.
\begin{align*}
\frac{\sin(8.00^\circ)}{c} & = \frac{\sin(26.20^\circ)}{2.3 \text{ mi}}.\\
\frac{c}{\sin(8.00^\circ)} & = \frac{2.3 \text{ mi}}{\sin(26.20^\circ)}.\\
c & = \frac{(2.3 \text{ mi})\sin(8.00^\circ)}{\sin(26.20^\circ)}\\
& \approx 0.7\underline{2}501447 \text{ mi}\\
& \approx 0.73 \text{ mi}\text{.}
\end{align*}
We now know the length of the hypotenuse of the right triangle whose leg is the desired aircraft altitude. This gives us angle-angle-side (first angle is the right angle). This means we can use the Law of Sines to calculate the height.
\begin{align*}
\frac{\sin(34.20^\circ)}{h} & = \frac{\sin(90^\circ)}{0.7\underline{2}501447 \text{ mi}}.\\
\frac{h}{\sin(34.20^\circ)} & = \frac{0.7\underline{2}501447 \text{ mi}}{\sin(90^\circ)}.\\
h & = \frac{(0.7\underline{2}501447 \text{ mi})\sin(34.20^\circ)}{\sin(90^\circ)}\\
& \approx 0.4\underline{0}75185828 \text{ mi}\\
& \approx 0.41 \text{ mi}\text{.}
\end{align*}
This is in miles. It will be easier to interpret in feet. The conversion ratio in Table 1.1.2 suggests we can multiply
\begin{equation*}
(0.4\underline{0}751862 \text{ mi}) \cdot \frac{5280 \text{ ft}}{1 \text{ mi}} \approx 2\underline{1}51.6983 \text{ ft} \approx 2200 \text{ ft}\text{.}
\end{equation*}
Exercises 7.3.5 Exercises
Solving Triangles.
Use the trigonometric laws to calculate side lengths and angles.
Applications.
Identify triangles in these applications and use trigonometric laws to answer these questions.
