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Lesson 24: Oblique Triangles and The Law of Sines & The Law of Cosines

### 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

• Develop the law of sines and use it to solve ASA and AAS triangles.
• Solve SSA triangles (the ambiguous case) using the law of sines.
• Use the law of sines to solve applications.

Topic. This lesson covers

Sections 7.1: Oblique Triangles and the Law of Sines, and

Section 7.2: The Law of Cosines.

WeBWorK. There are two WeBWorK assignments on today’s material:

LawOfSines

LawOfCosines

Lesson Notes.

Video Lesson.

Video Lesson 24 (based on Lecture 24 Notes)

### 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 $\arccos a$?

$\arccos a = \theta$

#### Warmup Question 2

Find $\arcsin 0.766$ in degrees.

$\arcsin 0.766=50^{\circ}$

### 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 23.

### 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 Oblique Triangles and The Law of Sines &

The 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

A description of the video

In the video you will see the following problems.

• Given a triangle whose angles are $40^{\circ}$, $60^{\circ}$ and $80^{\circ}$ where the sides opposite to $40^{\circ}$, $60^{\circ}$ and $80^{\circ}$ measure 5, $x$ and $y$, respectively, find $x$ and $y$.
• Given a triangle whose angles are $\theta$ and $95^{\circ}$, and the sides opposite to $\theta$ and $95^{\circ}$ measure 5 and $10$, respectively, find $\theta$.
• 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.

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:

LawOfSines

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 sines?
• 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 $b=250\;m$.

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.