# 5.3: Tangent Ratio

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

## Learning Objectives

- Identify and use the tangent ratio in a right triangle.
- Understand tangent ratios in special right triangles.

## The Tangent Ratio

The first ratio to examine when studying right triangles is the **tangent**.

The tangent of an angle is the *ratio* of the length of the **opposite** side to the length of the **adjacent** side. The **hypotenuse** is not involved in the tangent at all.

Recall that a *ratio* is the same as a *fraction,* so the tangent is the *fraction* of the opposite side over the adjacent side.

This means that tangent is: the ____________________ side divided by the ____________________ side.

Be sure when you find a **tangent** that you find the **opposite** and **adjacent** sides *relative to the angle in question.*

You must be careful that the **opposite** side is *across from* the angle you are taking the tangent of, and the **adjacent** side is *next to* that same angle!

For an acute angle measuring \begin{align*}\theta\end{align*}, we define:

\begin{align*}\tan \theta = \frac{opposite}{adjacent}\end{align*}

Like always, be sure to reduce the fraction in your final answer!

**Reading Check:**

1. *Fill in the blank: Another word for ratio is* _______________________.

2. *Which side of a right triangle is* *NOT**used in the tangent ratio?*

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

**Example 1**

*What are the tangents of* \begin{align*}\angle X \end{align*} *and* \begin{align*}\angle Y\end{align*} *in the triangle below?*

To find these ratios, first identify the sides **opposite** and **adjacent** to each angle:

For angle \begin{align*}X\end{align*}:

The side **opposite** angle \begin{align*}X\end{align*} is the segment ___________, which is _______ cm long.

The side **adjacent** to angle \begin{align*}X\end{align*} is the segment ___________, which is _______ cm long.

For angle \begin{align*}Y\end{align*}:

The side **opposite** angle \begin{align*}Y\end{align*} is the segment ___________, which is _______ cm long.

The side **adjacent** to angle \begin{align*}Y\end{align*} is the segment ___________, which is _______ cm long.

\begin{align*}\tan \angle X & =\frac{opposite}{adjacent}=\frac{5}{12}\\ \tan \angle Y & = \frac{opposite}{adjacent}=\frac{12}{5}\end{align*}

So, the tangent of\begin{align*}\angle X \end{align*} is \begin{align*}\frac{5}{12}\end{align*} and the tangent of \begin{align*}\angle Y \end{align*} is \begin{align*}\frac{12}{5}\end{align*}.

Notice that the tangent is different for different angles because of which sides are **opposite** and which sides are **adjacent** to each angle in the triangle.

It is common to write \begin{align*}\tan X\end{align*} instead of \begin{align*}\tan\angle X\end{align*}. In this text we will use both notations.

**Reading Check:**

*On the blank triangle above,*

1. *Label the right angle.*

2. *Label the triangle* \begin{align*}\Delta CAT\end{align*}, *where the right angle is at angle* \begin{align*}A\end{align*}.

3. *Label the hypotenuse.*

4. *On the side opposite angle* \begin{align*}C\end{align*}, *label “opposite* \begin{align*}\angle C\end{align*}”

5. *On the side adjacent to angle* \begin{align*}C\end{align*}, *label “adjacent* \begin{align*}\angle C\end{align*}”

6. *On the side opposite angle* \begin{align*}T\end{align*}, *label “opposite* \begin{align*}\angle T\end{align*}”

7. *On the side adjacent to angle* \begin{align*}T\end{align*}, *label “opposite* \begin{align*}\angle T\end{align*}”

*(Note: some sides will have more than one label!)*

## Tangents of Special Right Triangles

It may help you to learn some of the most common values for tangent ratios. The list below shows you values for angles in special right triangles.

Tangent \begin{align*}30^\circ\end{align*}: \begin{align*}\frac{1}{\sqrt{3}}\cdot \frac {\sqrt{3}}{\sqrt {3}} = \frac {\sqrt{3}}{3}\end{align*} this is approximately equal to (or \begin{align*}\approx\end{align*}) 0.577

Tangent \begin{align*}45^\circ\end{align*}: \begin{align*}\frac{1}{1} = 1\end{align*}

Tangent \begin{align*}60^\circ\end{align*} : \begin{align*}\frac{\sqrt{3}}{1} = \sqrt{3} \approx 1.732\end{align*}

Notice that you can derive these ratios from the \begin{align*}30^\circ - 60^\circ - 90^\circ\end{align*} special right triangle. We will see this on the following page.

The triangle below is labeled with the side lengths that correspond to a \begin{align*}30^\circ - 60^\circ - 90^\circ\end{align*} triangle, the shortest side being \begin{align*}x\end{align*} (as you saw in Unit 3):

If we start with the \begin{align*}30^\circ\end{align*} angle at the top of the triangle: the **opposite** side is \begin{align*}x\end{align*} and the **adjacent** side is \begin{align*}x\sqrt{3}\end{align*}.

Therefore, tangent of \begin{align*}30^\circ = \frac{x}{x\sqrt{3}} = \frac{1}{\sqrt{3}}\end{align*} or \begin{align*}\frac{\sqrt{3}}{3}\end{align*}

Next we will use the \begin{align*}60^\circ\end{align*} angle on the right of the triangle: the **opposite** side is \begin{align*}x\sqrt{3}\end{align*} and the **adjacent** side is \begin{align*}x\end{align*}.

Therefore, tangent of \begin{align*}60^\circ = \frac {x\sqrt{3}}{x} =\frac{\sqrt{3}}{1}\end{align*} or \begin{align*}\sqrt{3}\end{align*}

In order to figure out tangent of the \begin{align*}45^\circ\end{align*}, we must consider a right **isosceles** triangle, which has one angle that is \begin{align*}90^\circ\end{align*} and the other two that are the same measure (remember the definition of an isosceles triangle!)

This means that a *right isosceles triangle* has one angle that is _______ and the other two that equal _______.

Likewise, both *legs* of the isosceles triangle are the *same* length! When you take the tangent of the opposite leg over the adjacent leg, the value is 1.

You can use these ratios to identify angles in a triangle. Work backwards from the ratio. If the ratio equals one of these values, you can identify the measurement of the angle.

**Example 2**

*What is*\begin{align*}m\angle{J}\end{align*} *in the triangle below?*

Find the tangent of \begin{align*}\angle{J}\end{align*} and compare it to the values of tangent for special right triangles.

\begin{align*}\tan \ J & = \frac {opposite}{adjacent}\\ & = \frac{5}{5}\\ & = 1 \end{align*}

So, the tangent of \begin{align*}\angle {J}\end{align*} is 1. If you look in the list of tangent values, you can see that an angle that measures \begin{align*}45^\circ\end{align*} has a tangent of 1. So, \begin{align*}m\angle {J}=45^\circ\end{align*}.

(You may also notice that the triangle in this example is a right **isosceles** triangle, so the measure of both angles \begin{align*}J\end{align*} and \begin{align*}K\end{align*} must be \begin{align*}45^\circ\end{align*}.)

**Example 3**

*What is* \begin{align*}m\angle {Z}\end{align*} *in the triangle below?*

Find the tangent of \begin{align*}\angle{Z}\end{align*} and compare it to the values of tangent for special right triangles.

\begin{align*}tan \ Z & = \frac {opposite}{adjacent}\\ & = \frac{5.2}{3}\\ & = 1.7{\overline{3}}\end{align*}

So, the tangent of \begin{align*}\angle {Z}\end{align*} is about 1.73. If you look at the values of tangent for special triangles a few pages back, you can see that an angle that measures \begin{align*}60^\circ\end{align*} has a tangent of 1.732.So,\begin{align*}m \angle Z \approx 60^\circ\end{align*}.

Another interesting thing to notice in this example is that \begin{align*}\Delta XYZ\end{align*} is a \begin{align*}30^\circ - 60^\circ - 90^\circ\end{align*} triangle. You will remember from Unit 3 that this means that the sides of the triangle have a special relationship as well. You can use this fact to see that:

\begin{align*}XY = 5.2 \approx 3\sqrt{3}\end{align*}

and if \begin{align*}XZ = 3\end{align*} then of course \begin{align*}XY = 3\sqrt{3}!\end{align*}

**Reading Check (Challenge):**

Below is a \begin{align*}45^\circ - 45^\circ - 90^\circ\end{align*} triangle.

1. *Which side length is the hypotenuse?*

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

2. *Show your work to find the* *tangent**of a* \begin{align*}45^\circ\end{align*} *angle.* *(Notice that it does NOT matter which* \begin{align*}45^\circ\end{align*} *angle you choose!)*

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

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