This section covers:

# Review of Right Triangle Trig

We learned about **Right Triangle Trigonometry** here, where we could “solve” triangles to find missing pieces (angles or sides).

Here is a review of the basic trigonometric functions, shown with both the **SOHCAHTOA** and **Coordinate System** Methods. Note that the second set of three trig functions are just the reciprocals of the first three; this makes it a little easier!

Remember that the **sin** (cos, and so on) of an angle is just a number!

We use the **Law of Sines** and **Law of Cosines** to “solve” triangles (find missing angles and sides) when we do not have a right triangle (which is called an **oblique triangle**). This is a little more complicated, and we **have to know which angles and sides we do have to know which Law to use**, but it’s not too bad.

**Note that the Law of Sines can still be used to solve Right Triangles, using the 90° angle as one of the angles. It just turns out that the sin of 90° is 1; so it turns into a SOH case. **

# Law of Sines

The **Law of Sines **(or **Sine Rule**) provides a simple way to set up proportions to get other parts of a triangle that isn’t necessarily a right triangle.

We use the Law of Sines when we have the following parts of a triangle, as shown below: Angle, Angle, Side (**AAS**), Angle, Side, Angle (**ASA**), and Side, Side, Angle (**SSA**). Note that it won’t work when we only know the Side, Side, Side (**SSS**) or the Side, Angle, Side (**SAS**) pieces of a triangle. (Remember that these are “in a row” or adjacent parts of the triangle).

The only problem is that sometimes, with the **SSA** case, depending on what we know about the other sides and angles of the triangle, the triangle could actually have two different shapes (one acute and one obtuse). For these case, we have to account for both those shapes (so we’ll basically have two answers for the triangle, or maybe even no triangle can be formed). This is called the **Ambiguous Case**, and we’ll discuss it below **here**.

Once we know the formula for the **Law of Sines**, we can look at a triangle and see if we have enough information to “solve” it. “**Solving a triangle**” means finding any unknown sides and angles for that triangle (there should be six total for each individual triangle).

Note that we usually depict **angles** in **capital letters**, and the **sides directly across from them** in the same letter, but in **lower case**:

Let’s do some problems; let’s first use the **Law of Sines** to find the indicated side or angle.

Also remember that if we can’t solve it from what we have, and we have **two of the three angles**, we can obtain the third angle from Geometry (the sum of angles in a triangle is 180°).

Also note that the triangles aren’t typically **drawn to scale**, meaning the angles and side measurements don’t exactly match the pictures. When I draw the triangles, I typically put the **A** and **B** angles on the “ground”.

Also notice that if you try to draw the triangles to scale, you’ll see that **larger angles are opposite larger sides**, and **smaller angles are opposite smaller sides**. This could be a way you can check to see if you’re getting the correct answers.

And don’t forget to put your calculator in “**DEGREE” mode**.

Also, notice that I tend to round the **angle measurements** to a **tenth** of a degree, and the **side measurements** two decimal places (**hundredths**). If you want more accurate measurements when you start calculating the other sides and angles, you can use the **STO>** function in your calculator, or retype the long decimals you may get. To do this, use “**STO>X**” after seeing the value you want stored, and then “**X**” when you want to retrieve that value.

Here are some problems where we need to “solve” the triangle, using the Law of Sines. Again, solving the triangle means finding all the missing parts, both sides and angles.

## Law of Sines Ambiguous Case (**SSA**)

When we have the Side-Side-Angle (in a row) case (**SSA**), we could have **one**, **two**, or **no** triangles formed, and we have to do extra work to determine which situation we have.

In these cases, I always like to draw my triangle with the **known angle **on the bottom left, so I can see what’s going on. If the side directly across from this angle is **less than** the side touching this angle, we will probably have an ambiguous case (or may have no triangle that can be formed).

In this situation, if we get an error message in the calculator when trying to get the other angle using **Law of Sines**, there is **no triangle** that can be formed with the numbers given (and that’s the answer). If we get an answer other than 90°, we will have **two triangles** that can be formed, and the second angle is 180 minus the angle we just got (one will be acute and one will be obtuse). We can then solve for two different triangles. (The given two sides and one angle for the two triangles will be the same). If we do get 90° for the second angle, we have **one right triangle**.

Here’s an illustration of this:

Let’s do some problems, so it won’t seem so confusing. Solve for all possible triangles with the given conditions:

**Note:** We can also solve ambiguous case triangles using the Law of Cosines and a graphing calculator here).

Here’s another type of problem you might encounter when learning about the **Law of Sines Ambiguous Case**:

**Problem**:

Given a triangle ABC, what values of a would result in making two triangles if A = 40° and b = 10?

**Solution:**

Let’s first draw the triangle:

In order for there to be two triangles, **a** must be less than 10, but greater than the height of the triangle, which is 10sin40°, or 6.427. So **6.427 < a < 10**.

# Law of Cosines

The **Law of Cosines **(or **Cosine Rule**) again provides a simple way to set up proportions to get other parts of a triangle that isn’t necessarily a right triangle.

We use the Law of Cosines when we have the following parts of a triangle, as shown below: Side, Angle, Side (**SAS**), and Side, Side Side (**SSS**). It also will work for the Side, Side, Angle (**SSA**) case, and we will see that here, but the Law of Sines is usually taught with this case, because of the **Ambiguous Case**.

(Remember that these are “in a row” or adjacent parts of the triangle).

Once we know the formula for the **Law of Cosines**, we can look at a triangle and see if we have enough information to “solve” it. “**Solving a triangle**” means finding any unknown sides and angles for that triangle (there should be six total for each individual triangle).

Again, note that we usually depict **angles** in **capital letters**, and the **sides directly across from them** in the same letter, but in **lower case**:

Let’s do some problems; let’s first use the **Law of Cosines** to find the indicated side or angle.

Remember again that the triangles aren’t typically **drawn to scale**, meaning the angles and side measurements don’t exactly match the pictures. When I draw the triangles, I typically put the ** A** and

**angles on the “ground”.**

*B*Again notice that if you try to draw the triangles to scale, you’ll see that **larger angles are opposite larger sides**, and **smaller angles are opposite smaller sides**. This could be a way you can check to see if you’re getting the correct answers.

And don’t forget to put your calculator in “**DEGREE” mode**.

Now let’s solve for all possible triangles with the given conditions.

**Caution**: When using the **Law of Cosines** to solve the whole triangle (all angles and sides), particularly in the case of an obtuse triangle, you have to either finish solving the whole triangle using **Law of Cosines** (which is typically more difficult), or use the **Law of Sines** starting with the next **smallest angle** (the angle across from the smallest side) first. This is because of the problem with ambiguous cases with triangles.

## Law of Cosines Ambiguous Case (SSA)

Let’s solve an **SSA Ambiguous Case** Triangle using the **Law of Cosines** instead of the **Law of Sines**. We can do this fairly easily using a **graphing calculator**; in fact the calculator can actually tell us how many triangles we will get! We’ll use the same problem that we used earlier.

# Areas of Triangles

In Geometry we learned that we can get the area of triangles quite easily if we know the **base** of the triangle and the **altitude** (which is a line that is perpendicular to the base and extends up to the top of the triangle): \(\text{Area}\,\,=\frac{1}{2}bh\) or \(\text{Area}\,\,=\frac{{bh}}{2}\).

Now that we know trig, we get the area of a triangle without having to know the altitude if we know two sides, and the angle inside the two sides (the Side-Angle-Side or **SAS** case), or three sides of the triangle (Side-Side-Side, or **SSS** case).

And remember that many times if you don’t initially have these cases, you can solve the triangle using the **Law of Sines** or **Law of Cosines** to get these angles and sides.

Here are the two formulas; note that one uses the **sin** function, and the other (**Heron**’s formula) uses the sides only, with one half the sum of the sides being what is called a **semiperimeter**:

# Applications/Word Problems

Here are some examples of applications of the Law of Sines, Law of Cosines, and Area of Triangles. **One big hint on doing these word problems is to try to draw the diagram (if they don’t give you one) as much to scale as possible, so you can see if your answers make sense! **** **For example, draw the angles as close to the correct angle measurements and sides in the proportion of the numbers they give you.

**Problem:**

Ali and Brynn are standing 250 yards apart. Both girls site an airplane with angles of elevation 40° and 45°, respectively. How far from the plane is Ali?

**Solution:**

This is a good example how we might use the **Law of Sines** to get distances that are typically difficult to measure. This stuff is really used in “real life”!

**Problem:**

In a parallelogram ABCD, AD = 450 feet, AB = 240 feet, and diagonal AC = 290 feet. What is the measure of angle BCD?

**Solution:**

Sorry to spring some heavy **Geometry** on you, but you’ll see it’s not that bad. By definition, a **parallelogram** is a quadrilateral (four-sided figure with straight sides) that has opposite parallel sides, and it turns out that opposite sides are equal. Parallel means never crossing, like railroad tracks.

For a parallelogram ABCD, we need to draw A-B-C-D in a row (in any direction) around the figure.

So let’s draw the parallelogram and see how we can solve the problem. Note that we need to know from Geometry that adjacent angles in a parallelogram add up to 180°. These are called **Same Side Interior** angles.

**Problem:**

Three dogs are sitting in a kitchen and waiting to get their dog food. It turns out that Dog A is 4.5 feet from Dog B, and Dog C is 2.5 feet from Dog A, as shown in the diagram below (note that the three dogs’ positions form a right (90°) angle).

The angles formed at Dogs A and B to the dog food are 40° and 80°, respectively, as shown in the diagram.

How far is Dog C from the dog food?

**Solution:**

This one’s a little tricky, since we have more than one triangle. We’ll first work with the triangle where we have enough information, and then use a common side to solve the parts of the triangle we want.

We can start out using the triangle between Dogs A and B and the dog food, where we have an **ASA** case (**Law of Sines)**:

**Problem:**

A cruise ship travels at a **bearing** of 45° at 15 mph for 3 hours, and changes course to a bearing of 120°. It then travels 10 mph for 2 hours. Find the distance the ship is from its **original position** and also its** bearing** from the original position.** **

**Solution:**

(Note that this problem is actually solved more easily using **vectors** here in the **Introduction to Vectors** section.)

Let’s first talk about what it means to have a **bearing** of a certain degree, since this is typically used in navigation. First of all, think of north as going up (positive ** y** axis), south as going down (negative

**axis), east as going to the right (positive**

*y***axis), and west as going to the left (negative**

*x***axis).**

*x*Unless otherwise noted, **bearing** is the measure of the **clockwise angle** that starts **due north **or** on the positive y axis** (initial side) and terminates a certain number of degrees (terminal side) from that due north starting place. (This is also written, as in the case of a bearing of 40° as “40° east of north”, or “N40°E”).

**Note: **Sometimes, you’ll see a bearing that includes more directions, such as 70° west of north, also written as N70°W. In this case, the angle will start due north (straight up, or on the positive ** y** axis) and go counterclockwise 70° (because it’s going west, or to the left, instead of east). Similarly, a bearing of 50° south of east, or E50°S, would be an angle that starts due east (on the positive

**axis) and go clockwise 50° clockwise (towards the south, or down). Also, if you see a bearing of southwest, for example, the angle would be 45° south of west, or 225° clockwise from north, and so on.**

*x*And each time a boat or ship changes course, you have to draw another line to the north to map its new bearing.

Here are some bearing examples:

In this problem, we also have to remember that **Distance = rate x time**, since we are given rates and times and need to calculate distances. We also need to use some Geometry, so if you haven’t had it yet, you may want to skip this problem.

Here’s one more **bearing problem**:

**Problem:**

Joa is standing 100 feet from her friend Rachel. From Joa, the bearing to Rachel is E25°S. From Joa to another friend, Emily, the bearing is S30°E. The bearing from Rachel to Emily is S80°W. What is the distance from Emily to Rachel?

**Solution: **

Probably the most difficult part is to drawing a picture of the problem:

**Problem:**

Jill, a surveyor, needs to approximate the **area** of a piece of land. She walks the perimeter of the land and measures the side distances and one angle, as shown below. What is the area of the piece of land?

**Solution:**

**Understand these problems, and practice, practice, practice!**

Use the **Right Triangle Button** on the MathType keyboard to enter a problem, and then click on Submit (the arrow to the right of the problem) to solve the problem. You can also click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the **Mathway** site, where you can register for the **full version** (steps included) of the software. You can even get math worksheets.

You can also go to the **Mathway** site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!

On to** Polar Coordinates, Equations and Graphs** – you’re ready!

Thank-you so much for such a wonderful website!

You’re welcome! Thanks so much for visiting the site; this makes me want to keep writing and finish 🙂 Lisa

The airplane problem: you mean 45 degrees instead of 70 degrees?

YES! Thanks so much for finding this – don’t know what I was thinking. Please let me know if you see anything else that doesn’t look right on the site. Thanks again 🙂 Lisa

Hi, my name is Ruth . Please help me solve this problem

An aircraft flies a triangular course. The first leg is 200km on a bearing of 115° and the second leg is 150km on a bearing of 230°. How long is the third leg of the course and on what bearing must the aircraft fly?

Thanks for writing! Here’s what I got for this problem (see figure). Does this make sense? Lisa

Please help me to answer this. A and B are two stations 1 mile apart, and B is due east of A. When an airplane is due north of A its angles of elevation at A and B are 37 and 23 degrees respectively, finds its altitude at that time.

Thanks for writing! Here’s how I’d do this: it’s best to draw a picture: (I’m not sure how accurate they want your answers) Does this make sense? Lisa

The bases of a trapezoid are 22 and 12 respectively. The angles at the extremities of one base are 65 and 40 degrees respectively. Find the two legs.

Can you help me with this problem?

Thanks in advance!

Hi! I found the answer to this question here – hope that helps! Lisa

Hope you’ll help me with this…

The diagonals of a parallelogram are 7 inches and 9 inches respectively; they intersect at an angle of 52 degrees. Find the sides of a the parallelogram.

Here’s how I’d do this problem: I’d use Law of Cosines with an SAS case (dividing each of the diagonals by 2 to get the sides): a^2 = b^2 + c^2 – 2bccos A, or a^2 = (4.5)^2 + (3.5)^2 – 2(4.5)(3.5)cos(52), so a (side of the parallelogram) = 3.62 inches. THen you can do this again to get the other sides of the parallegram – I get 7.204 inches. Does that make sense? Let me know if you want me to send a picture. Lisa

Please help me answer this ma’am. A flagpole 25 feet tall stands on top of a building. From a point in the same horizontal plane with the base of the building the angles of elevation of the top and the bottom of the flagpole are 61 degrees 30′ and 56 degrees 20′ respectively. How high is the building? Thank you in advance.

Hi! I found this problem here Hope that helps! Lisa

Please help me answer this ma’am. A military observer notes two enemy batteries which subtend, at his observation, an angle of 40 degrees. the interval between the flash and the report of a gun is 5 seconds for one battery and 4 seconds for the other. If the velocity of sound is 1140 feet a second, how far apart are the batteries?

Hi! I found this problem here – hope that helps! Lisa

Can you send us picture of the drawing ma’am? Thank you. It would be a great help 🙂

Here’s a picture – hope that helps! Lisa

http://www.shelovesmath.com/wp-content/uploads/2015/09/IMG_71261.jpg

Hi! I would just like to ask help on how to illustrate the triangle from this problem:

A plane flew at 130 degrees from A to B and then at 225 degrees from B to C. If A is 538 miles from B and 807 miles from C, find the distance from B to C and the direction from A to C.

I just would like to know the illustration for the triangle here.

Thank you in advance!

Thanks for writing! Here’s how I’d draw this: Does that make sense? Lisa

Hi! will you please help me with this. thank you!

two ships leave the port at the same time. one ship sails on a course of 125 degrees at 18 knots while the other sails on a course 230 degrees at 24 knots. find after 3 hours the distance between the ships and the bearing from the first ship to the second.

thank you so much.

Thanks so much for writing. To get the distance between the two ships, I used the Law of Cosines: c^2 = a^2 + b^2 – 2abcos(C), or c^2 = 54^2 + 72^2 – 2*54*72Cos105; c = 100.56.

To get the bearing from ship A to B, I would use Law of Sines: sin(105)/100.56 = sin(x)/72. x = 43.76 degrees. To get this bearing, subtract 35 degrees, and then subtract from 270 to get 261.2 degrees. If you want me to send picture, let me know (and I’m not sure if I did the bearing stuff right 😉 Lisa

can you help me with this problem?

the bases of a trapezoid are 22 and 12 respectively. the angles at the extremities of one base are 65 degree and 40 degree respectively find the two legs. with drawing

Thanks for writing! The trick of doing this is to divide up the trapezoid into triangles and then use Law of Sines. Here is a solution I found:

http://www.jiskha.com/display.cgi?id=1328872726

Does this make sense? Can you draw a picture from this? Thanks, Lisa

Can someone please draw a picture for this problem:

Two sailboats leave a harbor at the same time. The first sails at 22mph in a direction 320 degrees. The second sails at 29mph in a direction at 200 degrees. Assuming that both boats maintain speed and heading after 2 hours, how far apart are the boats?

Thanks for writing and sorry it took so long to answer. Here’s a problem similar to this one – does it make sense? Look at number 9 here:

http://www.mathguy.us/BySubject/Trigonometry/Trigonometry_Semester_Review_Ch6_QandA.pdf

Lisa

Hi can you help me in this problema please 🙂 ? thank you 🙂

A triangular playground has sides of lenghts 475 feet, 595 feet, and 401 feet. what are the measures of the angles between the sides, to the nearest tenth of a degree?

Thanks for writing! For this problem, use the Law of Cosines, so to get the angle between the 475 and 595 side, 401^2 = 475^2 + 595^2 – 2*475*595cosC, C=42.2 degrees. Then you can do the same for one more side and find the last side by subtracting the first two from 180. Does that make sense? Lisa (I also found this problem

here).