Hi Everyone!

On this page you will find some material about Lesson 26. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

**Lesson 26: Solving Right Triangles & Applications of Static Trigonometry**

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 Outcomes.** (from Coburn and Herdlick’s Trigonometry book)

- Solve a right triangle given one angle and one side.
- Solve a right triangle given two sides.
- Solve applications involving angles of elevation and depression.
- Solve applications involving angles of rotation.
- Solve general applications of right triangles.

**Topic**. This lesson covers

Section 2.2: Solving Right Triangles, and

Section 2.3: Applications of Static Trigonometry.

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

SolvingRightTriangles

SolvingRightTriangles-InverseTrig

**Lesson Notes.**

**Video Lesson.**

Video Lesson 26 – part 1 (based on Lesson 26 Notes – part 1)

Video Lesson 26 – part 2 (based on Lesson 26 Notes – part 2)

### 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

What is the sum of the angles of a triangle?

#### Show Answer 1

$180^{\circ}$

#### Warmup Question 2

What is a right triangle?

#### Show Answer 2

It is a triangle that has an angle of $90^{\circ}$, that is, a right angle.

#### Warmup Question 3

What are the sides of a right triangle called?

#### Show Answer 3

Hypotenuse and legs.

### 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 ??

### 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 Solving Right Triangles & Applications of Static Trigonometry**

**Key Words.** Right angle, hypotenuse, leg, opposite leg, adjacent leg, Pythagorean Theorem, sine, cosine, tangent, cosecant, secant, cotangent, arcsine, arccosine, arctangent, solving a right triangle, special triangle, 30-60-90, 45-45-90, angle of depression and angle of elevation.

$\bigstar$ A **right angle** is an angle that measures $90^{\circ}$.

$\bigstar$ A **right triangle** is a triangle with a right angle.

$\bigstar$ The sum of the angles in a triangle is $180^{\circ}$.

$\bigstar$ **Sine**: $\sin \theta=\dfrac{\text{opposite leg}}{\text{hypotenuse}}$

$\bigstar$ **Pythagorean Theorem**: In a right triangle, if the legs measure $a$ and $b$ and the hypotenuse measures $c$, then

$$a^2+b^2=c^2.$$

$\bigstar$ **Cosine**: $\cos \theta=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}$

$\bigstar$ **Tangent**: $\tan\theta = \dfrac{\sin\theta}{\cos\theta}\text{ or } \dfrac{\text{opposite leg}}{\text{adjacent leg}}$

$\bigstar$ **Cosecant**: $\csc\theta = \dfrac{1}{\sin\theta}$

$\bigstar$ **Secant**: $\sec\theta = \dfrac{1}{\cos\theta}$

$\bigstar$ **Cotangent**: $\cot\theta = \dfrac{1}{\tan\theta}$

$\bigstar$ **Arcsine**: if $\sin\theta = a$, then $\arcsin(a) = \theta$

$\bigstar$ **Arccosine**: if $\cos\theta = a$, then $\arccos(a) = \theta$

$\bigstar$ **Arctangent**: if $\tan\theta = a$, then $\arctan(a) = \theta$

$\bigstar$ **Solving a right triangle** means to find the unknown angles and sides.

$\bigstar$ $30-60-90$ **Special Triangle**: This is a triangle whose angles are $30^\circ$, $60^\circ$ and $90^\circ$. This triangle is special, because the sides are in a special proportion. If the short leg (the opposite leg to $30^\circ$) is $x$, then

$$\text{the long leg (the opposite leg to } 60^\circ \text{) is }\sqrt 3x,$$

$$\text{the hypotenuse is } 2x.$$

$\bigstar$ $45-45-90$ **Special Triangle**: This is a triangle whose angles are $45^\circ$, $45^\circ$ and $90^\circ$. This triangle is special, because the sides are in a special proportion. If the legs are $x$, then

$$\text{the hypotenuse is } \sqrt 2x.$$

$\bigstar$ The angle made by the line of sight of an observer on the ground to a point above the horizontal is called the **angle of elevation**.

$\bigstar$ The angle made by the line of sight of an observer above to a point on the ground is called the **angle of depression**.

### 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 this video you will see the following problem:

A helicopter is flying 1,000 ft over a building. The pilot spots a person with an angle of depression $40^\circ$. How far is the person from the building?

### 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 right triangle.

(a picture of a right triangle taken from *Elementary College Geometry* by Henry Africk)

#### Show Answer 1

$\bigstar$ $$\cos 20^{\circ} = \dfrac{x}{10}$$

$$x = 10\cos20^{\circ}$$

$$x\approx 9.4$$

$\bigstar$ Let $y$ be the opposite side to the angle $20^{\circ}$. By using the Pythagorean Theorem, we obtain that

$$(9.4)^2 + y^2 = 10^2$$

$$88.36 + y^2 = 100$$

$$y^2=11.64$$

$$y\approx 3.4$$

$\bigstar$ The unknown angle $\hat{B}$ is

$$20^\circ+90^\circ+\hat B = 180^\circ$$

$$110^\circ+\hat B = 180^\circ$$

$$\hat B = 70^\circ$$

#### Try Question 2

Do not use a calculator in this question. Use diagrams to support your answers. The hypotenuse of a 45-45-90 triangle measures $5\sqrt 2$ cm.

(a) Find the length of its legs.

(b) Based on your answer in (a), find $\cos 45^{\circ}$, $\sin 45^{\circ}$ and $\tan 45^{\circ}$ in exact form.

#### Show Answer 2

(a) Both of its legs measure $5$ cm.

(b) $\cos45^{\circ}=\dfrac{adj}{hyp} = \dfrac{5}{5\sqrt 2} = \dfrac{1}{\sqrt 2} = \dfrac{\sqrt 2}{2}$

$\sin45^{\circ}=\dfrac{opp}{hyp} = \dfrac{5}{5\sqrt 2} = \dfrac{1}{\sqrt 2} = \dfrac{\sqrt 2}{2}$

$\tan45^{\circ}=\dfrac{opp}{adj} = \dfrac{5}{5} = 1$

#### Try Question 3

Solve the triangle ABC given that

\[\angle C=90^{\circ}, \qquad\qquad \angle B=40^{\circ}, \qquad\qquad b = 22\;\text{ft}.\]

Round to the nearest 100th.

#### Show Answer 3

We have

$$\angle A = 180-90-40 = 50^{\circ},$$

$$\sin 40^{\circ} = \dfrac{22}{c},$$

so

$$c=\dfrac{22}{\sin 40^{\circ}}\approx 34.23$$

$$a = \sqrt{c^2-b^2} = \sqrt{34.23^2-22^2} \approx 26.22$$

Therefore

$\bullet$ **angles:**

$\angle A = 50^{\circ}$

$\angle B = 40^{\circ}$

$\angle C = 90^{\circ}$

$\bullet$ **sides:**

$a = 26.22$

$b= 22$

$c= 34.23$

### 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:

SolvingRightTriangles

SolvingRightTriangles-InverseTrig

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

- What is the importance in drawing a picture for word problems?
- What is the relationship between an angle of depression and an angle of elevation? Give an example.
- How do we use our calculator to find an unknown angle in a right triangle if two sides are given?
- How is this related to finding the positive solution to the equation $x^2=5$ using our calculator?

$\bigstar$ Consider a 30-60-90 triangle with the longer leg measuring 9 inches.

(a) Find the length of the unknown sides.

(b) Find $\cos 60^{\circ}$, $\sin 60^{\circ}$ and $\tan 60^{\circ}$ in exact form using the above triangle.

#### Show Answer

(a) In a 30-60-90 triangle, the hypotenuse is $2x$ and the long leg is $\sqrt 3x$ where $x$ is the short leg. If the long leg is $9$ inches, we have that

\[\sqrt 3x=9 \quad\Longrightarrow\quad x = \dfrac{9}{\sqrt 3} = \dfrac{9\sqrt 3}{\sqrt 3\cdot \sqrt 3} = \dfrac{9\sqrt 3}{3} = 3\sqrt 3\]

So the length of the hypotenuse is $2x=2\cdot 3\sqrt 3 =6\sqrt 3$ inches, and the length of the short leg is $x=3\sqrt 3$ inches.

(b) We have that

\[\cos 60^{\circ} = \dfrac{\text{adjacent leg}}{\text{hypotenuse}} = \dfrac{3\sqrt{3}}{6\sqrt 3} = \dfrac{1}{2}. \]

\[\sin 60^{\circ} = \dfrac{\text{opposite leg}}{\text{hypotenuse}} = \dfrac{9}{6\sqrt 3} = \dfrac{\sqrt 3}{2}. \]

\[\tan 60^{\circ} = \dfrac{\text{opposite leg}}{\text{adjacent leg}} = \dfrac{9}{3\sqrt{3}} = \sqrt 3. \]

### 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.