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# 5.3: The Law of Sines

Created by: CK-12

## Learning Objectives

• Understand how both forms of the Law of Sines are obtained.
• Apply the Law of Sines when you know two angles and a non-included side and if you know two angles and the included side.
• Use the Law of Sines in real-world and applied problems.

We have learned about the Law of Cosines, which is a generalization of the Pythagorean Theorem for non-right triangles. We know that we can use the Law of Cosines when:

1. We know two sides of a triangle and the included angle (SAS) or
2. We know all three sides of the triangle (SSS)

But, what happens if the triangle we are working with doesn’t fit either of those scenarios? Here we introduce the Law of Sines.

The Law of Sines is a statement about the relationship between the sides and the angles in any triangle. While the Law of Sines will yield one correct answer in many situations, there are times when it is ambiguous, meaning that it can produce more than one answer. We will explore the ambiguity of the Law of Sines in the next section.

We can use the Law of Sines when:

1. We know two angles and a non-included side (AAS) or
2. We know two angles and the included side (ASA)

## Deriving the Law of Sines

$\triangle{ABC}$ contains altitude $CE$, which extends from $C$ and intersects $AB$. We will refer to the length of altitude $CE$ as $x$.

We know that $\sin A = \frac{x}{b}$ and $\sin B = \frac{x}{a}$, by the definition of sine. If we cross-multiply both equations and substitute, we will have the Law of Sines.

$& b (\sin A) = x \qquad \text{and} \qquad a (\sin B) = x \\& \qquad \searrow \qquad \qquad \qquad \qquad \swarrow \\& \qquad \qquad b (\sin A) = a (\sin B) \\& \qquad \frac{\sin A}{a} = \frac{\sin B}{b} \ \text{or}\ \frac{a}{\sin A} = \frac{b}{\sin B}$

Extending these ratios to angle $C$ and side $c$, we arrive at both forms of the Law of Sines:

$& \text{Form}\ 1: && \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \\& \text{(sines over sides)} \\& \text{Form}\ 2: && \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \\& \text{(sides over sines)}$

## AAS (Angle-Angle-Side)

One case where we can to use the Law of Sines is when we know two of the angles in a triangle and a non-included side (AAS).

Example 1: Using $\triangle{GMN}, \angle{G} = 42^\circ, \angle{N} = 73^\circ$ and $g = 12$. Find $n$.

Since we know two angles and one non-included side $(g)$, we can find the other non-included side $(n)$.

$\frac{\sin 73^\circ}{n} & = \frac{\sin 42^\circ}{12} \\n \sin 42^\circ & = 12 \sin 73^\circ \\n & = \frac{12 \sin 73^\circ}{\sin 42^\circ} \\n & \approx 17.15$

Example 2: Continuing on from Example 1, find $\angle{M}$ and $m$.

Solution: $\angle{M}$ is simply $180^\circ - 42^\circ - 73^\circ = 65^\circ$. To find side $m$, you can now use either the Law of Sines or Law of Cosines. Considering that the Law of Sines is a bit simpler and new, let’s use it. It does not matter which side and opposite angle you use in the ratio with $\angle{M}$ and $m$.

Option 1: $\angle{G}$ and $g$

$\frac{\sin 65^\circ}{m} & = \frac{\sin 42^\circ}{12} \\m \sin 42^\circ & = 12 \sin 65^\circ \\m & = \frac{12 \sin 65^\circ}{\sin 42^\circ} \\m & \approx 16.25$

Option 2: $\angle{N}$ and $n$

$\frac{\sin 65^\circ}{m} & = \frac{\sin 73^\circ}{17.15} \\m \sin 73^\circ & = 17.15 \sin 65^\circ \\m & = \frac{17.15 \sin 65^\circ}{\sin 73^\circ} \\m & \approx 16.25$

Example 3: A business group wants to build a golf course on a plot of land that was once a farm. The deed to the land is old and information about the land is incomplete. If $AB$ is 5382 feet, $BC$ is 3862 feet, $\angle{AEB}$ is $101^\circ,\angle{BDC}$ is $74^\circ,\angle{EAB}$ is $41^\circ$ and $\angle{DCB}$ is $32^\circ$, what are the lengths of the sides of each triangular piece of land? What is the total area of the land?

Solution: Before we can figure out the area of the land, we need to figure out the length of each side. In triangle $ABE$, we know two angles and a non-included side. This is the AAS case. First, we will find the third angle in triangle $ABE$ by using the Triangle Sum Theorem. Then, we can use the Law of Sines to find both $AE$ and $EB$.

$\angle{ABE} & = 180 - (41 + 101) = 38^\circ \\\frac{\sin 101}{5382} & = \frac{\sin 38}{AE} && \frac{\sin 101}{5382} = \frac{\sin 41}{EB} \\AE (\sin 101) & = 5382 (\sin 38) && EB (\sin 101) = 5382 (\sin 41) \\AE & = \frac{5382(\sin 38)}{\sin 101} && EB = \frac{5382(\sin 41)}{\sin 101} \\AE & = 3375.5\ feet && EB \approx 3597.0\ feet$

Next, we need to find the missing side lengths in triangle $DCB$. In this triangle, we again know two angles and a non-included side (AAS), which means we can use the Law of Sines. First, let’s find $\angle{DBC} = 180 - (74 + 32) = 74^\circ$. Since both $\angle{BDC}$ and $\angle{DBC}$ measure $74^\circ$, triangle $DCB$ is an isosceles triangle. This means that since $BC$ is 3862 feet, $DC$ is also 3862 feet. All we have left to find now is $DB$.

$\frac{\sin 74}{3862} & = \frac{\sin 32}{DB}\\DB (\sin 74) & = 3862 (\sin 32)\\DB & = \frac{3862(\sin 32)}{\sin 74}\\DB & \approx 2129.0\ feet$

Finally, we need to calculate the area of each triangle and then add the two areas together to get the total area. From the last section, we learned two area formulas, $K = \frac{1}{2} \ bc \sin A$ and Heron’s Formula. In this case, since we have enough information to use either formula, we will use $K = \frac{1}{2} \ bc \sin A$ since it is less computationally intense.

First, we will find the area of triangle $ABE$.

Triangle $ABE$:

$K & = \frac{1}{2}(3375.5)(5382)\sin 41 \\K & = 5,959,292.8\ ft^2$

Triangle $DBC$:

$K & = \frac{1}{2}(3862)(3862)\sin 32 \\K & = 3, 951,884.6\ ft^2$

The total area is $5,959,292.8 + 3,951,884.6 = 9,911,177.4\ ft^2$.

## ASA (Angle-Side-Angle)

The second case where we use the Law of Sines is when we know two angles in a triangle and the included side (ASA). For instance, in $\triangle{TRI}$:

$\angle{T},\angle{R}$, and $i$ are known

$\angle{T},\angle{I}$,and $r$ are known

$\angle{R},\angle{I}$, and $t$ are known

In this case, the Law of Sines allows us to find either of the non-included sides.

Example 4: (Use the picture above) In $\triangle{TRI},\angle{T} = 83^\circ, \angle{R} = 24^\circ$, and $i = 18.5$. Find the measure of $t$.

Solution: Since we know two angles and the included side, we can find either of the non-included sides using the Law of Sines. Since we already know two of the angles in the triangle, we can find the third angle using the fact that the sum of all of the angles in a triangle must equal $180^\circ$.

$\angle{I} & = 180 - (83 + 24) \\\angle{I} & = 180 - 107 \\\angle{I} & = 73^\circ$

Now that we know $\angle{I} = 73^\circ$, we can use the Law of Sines to find $t$.

$\frac{\sin 73}{18.5} & = \frac{\sin 83}{t} \\ t (\sin 73) & = 18.5 (\sin 83) \\t & = \frac{18.5(\sin 83)}{\sin 73} \\t & \approx 19.2$

Notice how we wait until the last step to input the values into the calculator. This is so our answer is as accurate as possible.

Example 5: In order to avoid a large and dangerous snowstorm on a flight from Chicago to Buffalo, pilot John starts out $27^\circ$ off of the normal flight path. After flying 412 miles in this direction, he turns the plane toward Buffalo. The angle formed by the first flight course and the second flight course is $88^\circ$. For the pilot, two issues are pressing:

1. What is the total distance of the modified flight path?
2. How much further did he travel than if he had stayed on course?

Solution, Part 1: In order to find the total distance of the modified flight path, we need to know side $x$. To find side $x$, we will need to use the Law of Sines. Since we know two angles and the included side, this is an ASA case. Remember that in the ASA case, we need to first find the third angle in the triangle.

$Missing Angle & = 180 - (27 + 88) = 65^\circ && \text{The sum of angles in a triangle is}\ 180 \\\frac{\sin 65}{412} & = \frac{\sin 27}{x} && \text{Law of Sines} \\x (\sin 65) & = 412 (\sin 27) && \text{Cross multiply} \\x & = \frac{412(\sin 27)}{\sin 65} && \text{Divide by sin}\ 65 \\x & \approx 206.4\ miles$

The total distance of the modified flight path is $412 + 206.4 = 618.4\ miles$.

Solution, Part 2: To find how much farther John had to travel, we need to know the distance of the original flight path, $y$. We can use the Law of Sines again to find $y$.

$\frac{\sin 65}{412} & = \frac{\sin 88}{y} && \text{Law of Sines} \\y (\sin 65) & = 412 (\sin 88) && \text{Cross multiply} \\y & = \frac{412 (\sin 88)}{\sin 65} && \text{Divide by}\ \sin 65 \\y & \approx 454.3\ miles$

John had to travel $618.4 - 454.3 = 164.1\ miles$ farther.

## Solving Triangles

The Law of Sines can be applied in many ways. Below are some examples of the different ways and situations to which we may apply the Law of Sines. In many ways, the Law of Sines is much easier to use than the Law of Cosines since there is much less computation involved.

Example 6: In the figure below, $\angle{C} = 22^\circ, BC = 12, DC = 14.3, \angle{BDA} = 65^\circ$, and $\angle{ABD} = 11^\circ$. Find $AB$.

Solution: In order to find $AB$, we need to know one side in $\triangle{ABD}$. In $\triangle{BCD}$, we know two sides and an angle, which means we can use the Law of Cosines to find $BD$. In this case, we will refer to side $BD$ as $c$.

$c^2 & = 12^2 + 14.3^2 - 2(12)(14.3) \cos 22 && \text{Law of Cosines} \\c^2 & \approx 30.28 \\ c & \approx 5.5$

Now that we know $BD \approx 5.5$, we can use the Law of Sines to find $AB$. In this case, we will refer to $AB$ as $x$.

$\angle{A} & = 180 - (11 + 65) = 104^\circ && \text{Triangle Sum Theorem} \\\frac{\sin 104}{5.5} & = \frac{\sin 65}{x} && \text{Law of Sines} \\x & = \frac{5.5 \sin 65}{\sin 104} && \text{Cross multiply and divide by}\ \sin 104 \\x & \approx 5.14$

Example 7: A group of forest rangers are hiking through Denali National Park towards Mt. McKinley, the tallest mountain in North America. From their campsite, they can see Mt. McKinley, and the angle of elevation from their campsite to the summit is $21^\circ$. They know that the slope of mountain forms a $127^\circ$ angle with ground and that the vertical height of Mt. McKinley is 20,320 feet. How far away is their campsite from the base of the mountain? If they can hike 2.9 miles in an hour, how long will it take them to get the base?

Solution: As you can see from the figure above, we have two triangles to deal with here: a right triangle $(\triangle{MON})$ and non-right triangle $(\triangle{MOU})$. In order to find the distance from the campsite to the base of the mountain, $y$, we first need to find one side of our non-right triangle, $\triangle{MOU}$. If we look at $\angle{M}$ in $\triangle{MNO}$, we can see that side $ON$ is our opposite side and side $x$ is our hypotenuse. Remember that the sine function is opposite/hypotenuse. Therefore we can find side $x$ using the sine function.

$\sin 21^\circ & = \frac{20320}{x} \\x \sin 21^\circ & = 20320 \\x & = \frac{20320}{\sin 21^\circ} \\x & \approx 56701.5$

Now that we know side $x$, we know two angles and the non-included side in $\triangle{MOU}$. We can use the Law of Sines to solve for side $y$. First, $\angle{MOU} = 180^\circ - 127^\circ - 21^\circ = 32^\circ$ by the Triangle Sum Theorem.

$\frac{\sin 127^\circ}{56701.5} & = \frac{\sin 32^\circ}{y} \\y \sin 127^\circ & = 56701.5 \sin 32^\circ \\y & = \frac{56701.5 \sin 32^\circ}{\sin 127^\circ} \\x & \approx 37623.2\ \text{or}\ 7.1\ miles$

If they can hike 2.9 miles per hour, then they will hike the 7.1 miles in 2.45 hours, or 2 hours and 27 minutes.

## Points to Consider

• Are there any situations where we might not be able to use the Law of Sines or the Law of Cosines?
• Considering what you already know about the sine function, is it possible for two angles to have the same sine? How might this affect using the Law of Sines to solve for an angle?
• By using both the Law of Sines and the Law of Cosines, it is possible to solve any triangle we are given?

## Review Questions

1. In the table below, you are given a figure and information known about that figure. Decide if each situation represents the AAS case or the ASA case.
Given Figure Case
a. $b = 16, A = 11.7^\circ, C = 23.8^\circ$
b. $e = 214.9, D = 39.7^\circ, E = 41.3^\circ$
c. $G = 22^\circ, I = 18^\circ, H = 140^\circ$
d. $k = 6.3, J = 16.2^\circ, L = 40.3^\circ$
e. $M = 31^\circ, O = 9^\circ, m = 15$
f. $Q = 127^\circ, R = 21.8^\circ, r = 3.62$
1. Even though ASA and AAS triangles represent two different cases of the Law of Sines, what do they both have in common?
2. Using the figures and the given information from the table above, find the following if possible:
1. side $a$
2. side $d$
3. side $i$
4. side $l$
5. side $o$
6. side $q$
3. In $\triangle{GHI}, \angle{I} = 21.3^\circ, \angle{H} = 62.1^\circ$, and $i = 108$. Find $g$ and $h$.
4. Use the Law of Sines to show that $\frac{a}{b} = \frac{\sin A}{\sin B}$ is true.
5. Use the Law of Sines, the Law of Cosines, and trigonometry functions to solve for $x$.
6. In order to avoid a storm, a pilot starts out $11^\circ$ off path. After he has flown 218 miles, he turns the plane toward his destination. The angle formed between his first path and his second path is $105^\circ$. If the plane traveled at an average speed of 495 miles per hour, how much longer did the modified flight take?
7. A delivery truck driver has three stops to make before she must return to the warehouse to pick up more packages. The warehouse, Stop A, and Stop B are all on First Street. Stop A is on the corner of First Street and Route 52, which intersect at a $41^\circ$ angle. Stop B is on the corner of First Street and Main Street, which intersect at a $103^\circ$ angle. Stop C is at the intersection of Main Street and Route 52. The driver knows that Stop A and Stop B are 12.3 miles apart and that the warehouse is 1.1 miles from Stop A. If she must be back to the warehouse by 10:00 a.m., travels at a speed of 45 MPH, and takes 2 minutes to deliver each package, at what time must she leave?

1. ASA
2. AAS
3. neither
4. ASA
5. AAS
6. AAS
1. Student answers will vary but they should notice that in both cases you know or can find an angle and the side across from it.
1. $\frac{\sin 11.7^\circ}{a} = \frac{\sin 144.5^\circ}{16}, a = 5.6$
2. $\frac{\sin 41.3^\circ}{214.9} = \frac{\sin 39.7^\circ}{d}, d = 208.0$
3. not enough information
4. $\frac{\sin 40.3^\circ}{l} = \frac{\sin 123.5^\circ}{6.3}, l = 4.9$
5. $\frac{\sin 9^\circ}{o} = \frac{\sin 31^\circ}{15}, o = 4.6$
6. $\frac{\sin 127^\circ}{q} = \frac{\sin 21.8^\circ}{3.62}, q = 7.8$
2. $\angle{G} = 180^\circ - 62.1^\circ - 21.3^\circ = 96.6^\circ$ $\frac{\sin 96.6^\circ}{g} = \frac{\sin 21.3^\circ}{108}, g = 295.3 && \frac{\sin 62.1^\circ}{h} = \frac{\sin 21.3^\circ}{108}, h = 262.8$
3. $\frac{\sin A}{a} & = \frac{\sin B}{b} && \text{Law of Sines} \\a (\sin B) & = b (\sin A) && \text{Cross multiply} \\\frac{a}{b} & = \frac{\sin A}{\sin B} && \text{Divide by}\ b(\sin B)$
1. $\tan 54^\circ = \frac{h}{7.15} \rightarrow h = 9.8, \cos 67^\circ = \frac{9.8}{x} \rightarrow x = 25.2$
2. The angle we are finding is the one at the far left side of the triangle.

$8.9^2 = 11.2^2 + 12.6^2 - 2 \cdot 11.2 \cdot 12.6 \cos A \rightarrow A = 43.4^\circ, \frac{\sin 43.4^\circ}{x} = \frac{\sin 31}{11.2} \rightarrow x = 14.9.$

4. First we need to find the other two sides in the triangle. $\frac{\sin 64^\circ}{218} = \frac{\sin 11^\circ}{x} = \frac{\sin 105^\circ}{y}, x = 46.3, y = 234.3$, where $y$ is the length of the original fight plan. The modified flight plan is $218 + 46.3 = 264.3$. Dividing both by 495 mi/hr, we get 32 min (modified) and 28.4 min (original). Therefore, the modified flight plan is 3.6 minutes longer.
5. First, we need to find the distance between Stop B (B) and Stop C (C). $\frac{\sin 36^\circ}{12.3} = \frac{\sin 41^\circ}{B} = \frac{\sin 103^\circ}{C} B = 13.7, C = 20.4$. The total length of her route is $1.1+12.3+13.7+20.4+1.1 = 48.6\ miles$. Dividing this by 45 mi/hr, we get that it will take her 1.08 hours, or 64.8 minutes, of actual driving time. In addition to the driving time, it will take her 6 minutes (three stops at 2 minutes per stop) to deliver the three packages, for a total roundtrip time of 70.8 minutes. Subtracting this 70.8 minutes from 10:00 am, she will need to leave by 8:49 am.

Feb 23, 2012

Aug 21, 2014