# 7.8: Using Trigonometric Ratios

**At Grade**Created by: CK-12

## Introduction

*The Ramp Dilemma (Part 2)*

**Here is the problem from the last lesson. Now that you know the math needed, think about this problem and the sine ratio.**

Mr. Watson’s homeroom decided to do a community service repairing the ramp outside the shed. The fresh coat of paint shone in the bright sunlight and Mr. Watson walked across the grass with all of his students to look at the ramp outside the door of the shed.

“Has that always been there?” asked Dan.

“No, in fact it was just brought out here yesterday,” Mr. Watson explained.

“Well if it’s brand new, then why do we need to fix it?” Emily asked.

“Because it doesn’t fit neatly under the door.”

Sure enough, the students looked and could see that the back of the ramp was too tall and the students would need to fix the back to make it fit beneath the doorway. Fixing this ramp would help everyone because it would make it easy to push or pull the cart with the athletic equipment out onto the field and since the equipment was used by local kid’s teams as well as the school, this was a definite way to give back to the community.

“What do we need to do?” Dan asked.

They looked at the ramp. Mr. Watson drew the following sketch.

“Well that’s not very helpful,” Dan commented.

“Sure it is,” Emily said.

**This lesson is about using trigonometric ratios. By the end of the lesson, you will know how to figure out the height of the ramp.**

*What You Will Learn*

By the end of this lesson, you will be able to complete the following skills.

- Use calculator or tables to find sine, cosine and tangent ratios for each of the acute angles of right triangles with given side lengths.
- Determine and use the sine ratio to find the length of an opposite side of a right triangle, given the angle and the hypotenuse.
- Determine and use the cosine ratio to find the length of an adjacent side of a right triangle, given the angle and hypotenuse.
- Determine and use the tangent ratio to find the height of a right triangle, given the angle and the base.
- Use calculators or tables to solve real – world problems involving right triangles and trigonometric ratios.

*Teaching Time*

I. **Use Calculator or Tables to Find Sine, Cosine and Tangent Ratios for Each of the Acute Angles of Right Triangles with Given Side Lengths**

Now that you have a basic understanding of trigonometric ratios and how they are formed, it is time to learn more about how they can be used. **You now know that a trigonometric ratio for a specific angle will remain constant no matter how large or small the triangle is. The idea is that the sides will always be in proportion to each other. So, if you know the measure of an angle (and can therefore identify the value of a trigonometric ratio) and the value of one side, you can use trigonometry to calculate the lengths of other sides.** The trick is to use good algebra technique, and make sure that every time you set up a ratio, you are putting the values and variables in the correct places.

**You can find trigonometric ratios by using your calculator.**

You understand trigonometric ratios and have had a chance to practice reading specific values out of a table. **You can find the ratio for any trigonometric value using your calculator. Take a moment to locate the buttons for sine, cosine, and tangent on the calculator. Keep in mind that usually, sine is abbreviated as** *sin***, cosine is usually abbreviated as** *cos***, and tangent is usually abbreviated as** ** tan.** Press the key of the ratio you want to find, and enter the angle in question. If you hit enter, or calculate, the calculator will show you the value of that specific ratio.

Example

*Use your calculator to find the values of the following ratios.*

\begin{align*}\text{sine} \ 47^{\circ}\end{align*}

\begin{align*}\text{cosine} \ 23^{\circ}\end{align*}

\begin{align*}\text{tangent} \ 82^{\circ}\end{align*}

**You can find the values for each ratio using your calculator. When dealing with large decimals values, it is usually best to round the numbers to the nearest thousandth. It gives you a reasonably accurate value without being too long of a number to work with.**

**The sine of \begin{align*}47^{\circ}\end{align*} is 0.73135370161917..., or about 0.731.**

**The cosine of \begin{align*}23^{\circ}\end{align*} is 0.92050485345244..., or about 0.921.**

**The tangent of \begin{align*}82^{\circ}\end{align*} is 7.11536972238419..., or about 7.115.**

*Notice that because we are using a calculator, that we don’t need to know the side lengths. The calculator figures out the trigonometric ratio based on proportions and the given angle. Figuring out sine, cosine and tangents using a calculator is as easy as pressing a button!*

II. **Determine and Use the Sine Ratio to Find the Length of an Opposite Side of a Right Triangle, Given the Angle and the Hypotenuse**

You have already learned how to write and determine the sine ratio, now we are going to use what we have learned in problem solving.

As you recall, the ratio of sine is \begin{align*}\frac{opposite}{hypotenuse}\end{align*}. If you know the sine value of the angle in question, and the length of the hypotenuse, you can find the measure of the opposite side. Look at the algebraic manipulation below.

\begin{align*}\sin e \angle X&=\frac{opposite}{hypotenuse}\\ \sin e \angle X \times hypotenuse&=\frac{opposite}{hypotenuse} \times hypotenuse\\ \sin e \angle X \times hypotenuse&=opposite\end{align*}

**If you multiply the sine of any angle \begin{align*}X\end{align*} and the length of the hypotenuse, the result is the length of the opposite side.**

*Write this statement down in your notebook. Be sure that you write that it is connected to the sine ratio.*

Now let’s apply this to an example.

Example

*What is the length of side \begin{align*}AC\end{align*} in the triangle below?*

**Use the following equation to find the length of the side opposite angle \begin{align*}B\end{align*}. Notice that we are going to multiply the sine of Angle \begin{align*}B\end{align*} by the length of the hypotenuse to find the other side. Therefore, you will have to find the sine of angle \begin{align*}B\end{align*} first. Then you multiply that times the length of the hypotenuse. Notice that we are finding the length of the opposite side with sine.**

\begin{align*} \sin e \angle B \times hypotenuse&=opposite\\ \sin e 14.5 \times 5&=opposite\\ 0.250 \times 5&=opposite\\ 1.25 &=opposite\end{align*}

**The length of side \begin{align*}AC\end{align*} is 1.25 units.**

We can apply a similar rule when we work with cosines. Let’s take a look.

III. **Determine and Use the Cosine Ratio to Find the Length of an Adjacent Side of a Right Triangle, given the Angle and the Hypotenuse**

Now as we work with cosines, we will be using given information to find the length of the adjacent side of a right triangle. Remember that the adjacent side is the side next to the angle that we are working with. As you recall, the ratio of cosine is \begin{align*}\frac{adjacent}{hypotenuse}\end{align*}. If you know the cosine value of the angle in question, and the length of the hypotenuse, you can find the measure of the adjacent side. Look at the algebraic manipulation below.

\begin{align*}\cos ine \angle X&=\frac{adjacent}{hypotenuse}\\ \cos ine \angle X \times hypotenuse&=\frac{adjacent}{hypotenuse} \times hypotenuse\\ \cos ine \angle X \times hypotenuse&=adjacent\end{align*}

**If you multiply the cosine of any angle \begin{align*}X\end{align*} and the length of the hypotenuse, the result is the length of the adjacent side.**

*Write this statement in your notebook. Be sure to include that it is for cosines.*

Example

*What is the length of side \begin{align*}BC\end{align*} in the triangle below?*

**Use the following equation to find the length of the side adjacent to angle \begin{align*}B\end{align*}. Notice that to find the length of the adjacent side that you will first need to find the cosine for angle \begin{align*}B\end{align*}. Then you can multiply that answer with the length of the hypotenuse. This will give you the measurement of the side next to or adjacent to the angle.**

\begin{align*}\cos ine \angle B \times hypotenuse&=adjacent\\ \cos ine 14.5 \times 5&=adjacent\\ 0.968 \times 5 &=adjacent\\ 4.84 &=adjacent\end{align*}

**The length of side \begin{align*}BC\end{align*} is 4.84 units.**

Finally, let’s look at tangents.

IV. **Determine and Use the Tangent Ratio to Find the Height of a Right Triangle, Given the Angle and the Base**

Now we can use the tangent ratio to find the length of the height of a right triangle. We can see that the height in a right triangle looks like side \begin{align*}a\end{align*}. This is because of the type of triangle that a right triangle is. As you recall, the ratio of tangent is \begin{align*}\frac{opposite}{adjacent}\end{align*}. If you know the tangent value of the angle in question, and the length of the adjacent side, you can find the measure of the opposite side. Look at the algebraic manipulation below.

\begin{align*}\tan gent \angle X&=\frac{opposite}{adjacent}\\ \tan gent \angle X \times adjacent &= \frac{opposite}{adjacent} \times adjacent\\ \tan gent \angle X \times adjacent &=opposite\end{align*}

**If you multiply the tangent of any angle \begin{align*}X\end{align*} and the length of the adjacent side, the result is the length of the opposite side.**

*Write this statement in your notebook. Be sure that it is connected to tangents.*

Example

*What is the length of side \begin{align*}QR\end{align*} in the triangle below?*

**Use the following equation to find the length of the side opposite angle \begin{align*}P\end{align*}. Notice that to find the measure of the opposite side we need to first find the tangent of the angle we are working with. Then we can take that measure and multiply it with the adjacent side.**

\begin{align*}\tan gent \angle P \times adjacent &=opposite\\ \tan gent \ 72 \times 8 &=opposite\\ 3.078 \times 8 &= opposite\\ 24.624 &= opposite\end{align*}

**The length of side \begin{align*}QR\end{align*} is 24.624 units.**

Now let’s look at how we can apply this information to real – life problems.

V. **Use Calculators or Tables to Solve Real – World Problems Involving Right Triangles and Trigonometric Ratios**

All of the ratios that you have been using are connected to Trigonometry. We can use trigonometry in real – world situations to solve real problems. It is important to think of trigonometry as a tool you have to solve different types of problems. Just as you know how to use the Pythagorean to find the third side of a right triangle when you know two, **use trigonometry to find missing sides when you know specific angle measurements in a triangle.** As you are given values in a problem, either create or add to a diagram to represent the information in the clearest possible way.

*Write this statement down in your notebook.*

Now let’s look at an example.

Example

*Craig launched a model rocket and wanted to calculate how high it went. He stood 10 feet away from the launch site and used a tool to calculate the angle between the ground and his line of sight to the rocket at its highest point. His data is shown in the diagram below.*

*Calculate the greatest height the model rocket reached.*

**In this problem, you have a lot of information, but the only important data is the shape of the triangle, the length of the base, and the measurement of the angle. The base is the side adjacent to where Craig is standing, and you want to find the value of the side opposite where Craig was standing. For these purposes, the hypotenuse of the triangle is irrelevant. Since you have the angle, and the adjacent side, you can find a tangent to discover the height of the triangle.**

\begin{align*}\tan gent &=\frac{opposite}{adjacent}\\ \tan 65^{\circ} &= \frac{opposite}{10 \ feet}\\ 2.145 &=\frac{opposite}{10 \ feet}\\ 2.145 \times 10 \ feet &= \frac{opposite}{10 \ feet} \times 10 \ feet\\ 21.45 \ feet &=opposite\end{align*}

**Craig’s rocket reached it’s highest height at 21.45 feet, almost 21 and \begin{align*}\frac{1}{2}\end{align*} feet high!**

## Real-Life Example Completed

*The Ramp Dilemma (Part 2)*

**Here is the original problem once again, be sure to reread it one last time.**

Mr. Watson’s homeroom decided to do a community service repairing the ramp outside the shed. Tache fresh coat of paint shone in the bright sunlight and Mr. Watson walked across the grass with all of his students to look at the ramp outside the door of the shed.

“Has that always been there?” asked Dan.

“No, in fact it was just brought out here yesterday,” Mr. Watson explained.

“Well if it’s brand new, then why do we need to fix it?” Emily asked.

“Because it doesn’t fit neatly under the door.”

Sure enough, the students looked and could see that the back of the ramp was too tall and the students would need to fix the back to make it fit beneath the doorway. Fixing this ramp would help everyone because it would make it easy to push or pull the cart with the athletic equipment out onto the field and since the equipment was used by local kid’s teams as well as the school, this was a definite way to give back to the community.

“What do we need to do?” Dan asked.

They looked at the ramp. Mr. Watson drew the following sketch.

“Well that’s not very helpful,” Dan commented.

“Sure it is,” Emily said.

*Now apply the sine ratio and figure out the height of the ramp.*

*Solution to Real – Life Example*

**First, we take the measurements and use the sine ratio.**

\begin{align*}\text{Sine} \ 15^{\circ} = \frac{Opposite}{Hypotenuse}\end{align*}

**The opposite in this example is the missing side. We use \begin{align*}x\end{align*} to represent this unknown measure. This is the measure that we are searching for.**

\begin{align*}\text{Sine} \ 15^{\circ} = \frac{opposite}{hypotenuse}=\frac{x}{4 \ ft}\end{align*}

**Now we can multiply both sides by 4 to find the measure.**

\begin{align*}4 \ \text{Sine} \ 15 &= x\\ 2.6 & = x\end{align*}

**The height of the ramp is 2.6 feet**

## Vocabulary

Here are the vocabulary words that are found in this lesson.

- Sine
- a ratio between the opposite side and the hypotenuse of a given angle.

- Cosine
- a ratio between the adjacent side and the hypotenuse of a given angle.

- Tangent
- a ratio between the opposite side and the adjacent side of a given angle.

- Trigonometric Ratio
- used to find missing side lengths of right triangles when angle measures have been given.

## Time to Practice

Directions: Answer each question about trigonometric ratios.

- True or false. There are three trigonometric ratios.
- True or false. The sine ratio is the opposite compared to the adjacent.
- True or false. The tangent ratio is the opposite compared to the adjacent.
- True or false. The sine ratio is the opposite compared to the hypotenuse.
- True or false. The cosine ratio is the adjacent compared to the hypotenuse.

Directions: Use the information given and what you have learned about trigonometric ratios to figure out the measure of each missing side.

- Tangent angle \begin{align*}A \ 45^{\circ}\end{align*}, adjacent side length – 5, what is the length of the opposite side?
- Tangent angle \begin{align*}B \ 12^{\circ}\end{align*}, adjacent side length – 6, what is the length of the opposite side?
- Tangent angle \begin{align*}C \ 20^{\circ}\end{align*}, adjacent side length – 15, what is the length of the opposite side?
- Sine angle \begin{align*}D \ 2^{\circ}\end{align*}, hypotenuse – 12, what is the length of the opposite side?
- Sine angle \begin{align*}E \ 65^{\circ}\end{align*}, hypotenuse – 8, what is the length of the opposite side?
- Sine angle \begin{align*}F \ 45^{\circ}\end{align*}, hypotenuse – 2, what is the length of the opposite side?

Directions: Answer each of the following questions.

- What is the tangent ratio?
- What is the cosine ratio?
- What is the sine ratio?

Directions: Draw a diagram for each problem and then solve each problem using trigonometric ratios. Each problem has two parts, a diagram and an answer.

- Mary wants to find out how tall a tree is. She knows if she measures the angle to from the ground to the tip of the tree 8 feet away, it is \begin{align*}85^{\circ}\end{align*}. How tall is the tree?
- Barry ties one end of string to the top of his apartment building, drops it to the ground, and then pulls it out from the building. If the angle the string makes with the ground is \begin{align*}75^{\circ}\end{align*} and the length of the string is 125 feet, how tall is the building?
- A suspension bridge has a large metal pipe connecting the roadway to the top of the tower. The angle the pipe makes with the road is \begin{align*}40^{\circ}\end{align*} and the length of the pipe is 75 yards. What is the distance between the place where the roadway and pipe intersect and the base of the tower?