Hi Everyone!
On this page you will find some material about Lesson 33. Read through the material below, watch the videos, and follow up with your instructor if you have questions.
Lesson 33: Law of Cosines
Table of Contents
Resources
In this section you will find some important information about the specific resources related to this lesson:
- the learning outcomes,
- the section in the textbook,
- the WeBWorK homework sets,
- a link to the pdf of the lesson notes,
- a link to a video lesson.
Learning Outcome (from Coburn and Herdlick’s Trigonometry textbook)
- Apply the law of cosines when two sides and an included angle are known (SAS).
- Apply the law of cosines when three sides are known (SSS).
- Solve applications using the law of cosines.
Topic. This lesson covers
Section 7.2: The Law of Cosines.
WeBWorK. There is one WeBWorK assignment on today’s material:
LawOfCosines
Lesson Notes.
These notes are used in Lessons 32 and 33. Today’s lesson is on pages 3 (SSS), 4 and 5.
Video Lesson.
Video Lesson 33 (based on Lesson 33 Notes)
This video is used in Lessons 32 and 33. For today’s lesson, watch from [8:02] to {13:05].
Warmup Questions
These are questions on fundamental concepts that you need to know before you can embark on this lesson. Don’t skip them! Take your time to do them, and check your answer by clicking on the “Show Answer” tab.
Warmup Question 1
If $\cos \theta = a$, what is $\arccos a$?
Show Answer 1
$\arccos a = \theta$
Warmup Question 2
Find $\arcsin 0.766$ in degrees.
Show Answer 2
$\arcsin 0.766=50^{\circ}$
Review
If you are not comfortable with the Warmup Questions, don’t give up! Click on the indicated lesson for a quick catchup. A brief review will help you boost your confidence to start the new lesson, and that’s perfectly fine.
Need a review? Check Lesson 32.
Quick Intro
This is like a mini-lesson with an overview of the main objects of study. It will often contain a list of key words, definitions and properties – all that is new in this lesson. We will use this opportunity to make connections with other concepts. It can be also used as a review of the lesson.
A Quick Intro to Law of Cosines
Key Words. SSS, SAS, AAS, ASA, Oblique triangle, solving a triangle, law of sines, law of cosines
$\bigstar$ We denote the sides of a tringle $\Delta ABC$ by $a$, $b$ and $c$ as the sides opposite to the angles $A$, $B$ and $C$, respectively.
$\bullet$ SSS means that the three sides are known.
$\bullet$ SAS means that two sides and the adjacent angle are known.
$\bullet$ AAS means that two angles and one side (not between the two angles) are known.
$\bullet$ ASA means that two angles and the adjacent side are known.
$\bigstar$ Solving a triangle means to find the unknown sides and angles.
$\bigstar$ A triangle that does not have a right angle is called oblique.
$\bigstar$ In a right triangle, you use the trig ratios to solve it. Otherwise, the triangle is oblique in which case consider:
$\bigstar$ Law of sines (for ASA/AAS triangles)
\[\dfrac{a}{\sin A} = \dfrac{b}{\sin B}=\dfrac{c}{\sin C}.\]
$\bigstar$ Law of cosines (for SAS/SSS triangles)
\[a^2=b^2+c^2-2bc\cos A\]
\[b^2=a^2+c^2-2ac\cos B\]
\[c^2=a^2+b^2-2ab\cos C\]
Video Lesson
Many times the mini-lesson will not be enough for you to start working on the problems. You need to see someone explaining the material to you. In the video you will find a variety of examples, solved step-by-step – starting from a simple one to a more complex one. Feel free to play them as many times as you need. Pause, rewind, replay, stop… follow your pace!
Video Lesson
This video is used in Lessons 32 and 33. For today’s lesson, watch from [8:02] to {13:05].
A description of the video
In the video you will see the following problems.
- How to solve an SSS triangle.
- Given a triangle whose one of the angles is $\theta$ and $95^{\circ}$, the side opposite to $\theta$ measures 3, and the two other sides measure 5 and 6, find $\theta$.
Try Questions
Now that you have read the material and watched the video, it is your turn to put in practice what you have learned. We encourage you to try the Try Questions on your own. When you are done, click on the “Show answer” tab to see if you got the correct answer.
Try Question 1
Solve the triangle having the following properties: side $b$ = 385 m, $\angle C = 67^{\circ}$, and side $a$ =490 m.
Show Answer 1
By the law of cosines,
$$c^2=a^2+b^2-2ab\cos C$$
$$c^2 = (490)^2+(385)^2-2\cdot 490\cdot 385\cdot\cos 67^{\circ}$$
$$c^2 \approx 240902.1452$$
$$c\approx 490.82$$
By the law of sines,
$$\dfrac{\sin A}{a}=\dfrac{\sin C}{c}$$
$$\dfrac{\sin A}{490} = \dfrac{\sin 67^{\circ}}{490.82}$$
$$\sin A =\dfrac{490\cdot \sin 67^{\circ}}{490.82} $$
$$\sin A\approx 0.91896699 $$
$$A\approx \arcsin (0.91896699)\approx 66.77 $$
Hence $B= 180-A-C \approx 180 – 66.77 – 67 =46.23$.
$\bullet$ sides:
$a=490$
$b=385$
$c\approx 490.82$
$\bullet$ angles:
$A\approx66.77^{\circ}$
$B\approx 46.23^{\circ}$
$C=67^{\circ}$
WeBWorK
You should now be ready to start working on the WeBWorK problems. Doing the homework is an essential part of learning. It will help you practice the lesson and reinforce your knowledge.
WeBWork
It is time to do the homework on WeBWork:
LawOfCosines
When you are done, come back to this page for the Exit Questions.
Exit Questions
After doing the WeBWorK problems, come back to this page. The Exit Questions include vocabulary checking and conceptual questions. Knowing the vocabulary accurately is important for us to communicate. You will also find one last problem. All these questions will give you an idea as to whether or not you have mastered the material. Remember: the “Show Answer” tab is there for you to check your work!
Exit Questions
- Why is there not an SSA theorem?
- When can you use the law of cosines?
- Why is it better to use more accuracy or exact answers in early parts of solving triangles?
$\bigstar$ Solve the triangle $\triangle ABC$ for which $\angle A=31^{\circ}$, $c=207\;m$, and $b=250\;m$.
Show Answer
By the law of cosines,
$$a^2=b^2+c^2-2bc\cos A$$
$$a^2 =(250)^2+(207)^2-2\cdot 250\cdot 207\cdot\cos 31^{\circ}$$
$$a^2 \approx 16632.18$$
$$a\approx 128.97 $$
By the law of sines,
$$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}$$
$$\dfrac{\sin 31^{\circ}}{128.97} = \dfrac{\sin B}{250}$$
$$\sin B = 250\cdot \dfrac{\sin 31^{\circ}}{128.97} $$
$$\sin B\approx 1$$
$$B\approx 90 $$
Hence $C= 180-A-B \approx 180 – 31 – 90 \approx 59$.
$\bullet$ sides:
$a\approx 128.97$
$b=250$
$c=207$
$\bullet$ angles:
$A=31^{\circ}$
$B\approx 90^{\circ}$
$C \approx 59^{\circ}$
Need more help?
Don’t wait too long to do the following.
- Watch the additional video resources.
- Talk to your instructor.
- Form a study group.
- Visit a tutor. For more information, check the tutoring page.
