# 1.10: Pythagorean Theorem for Solving Right Triangles

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

**Practice**Pythagorean Theorem for Solving Right Triangles

You notice that the rope attached to the tetherball is 1 meter long, and that the angle between the rope and the pole is \begin{align*}35^\circ\end{align*}

### Pythagorean Theorem

You can use your knowledge of the Pythagorean Theorem and the six trigonometric functions to solve a right triangle. Because a right triangle is a triangle with a 90 degree angle, solving a right triangle requires that you find the measures of one or both of the other angles. How you solve for these other angles, as well as the lengths of the triangle's sides, will depend on how much information is given.

Let's take a look at some problems involving the Pythagorean Theorem.

1. Solve the triangle shown below.

** **

We need to find the lengths of all sides and the measures of all angles. In this triangle, two of the three sides are given. We can find the length of the third side using the Pythagorean Theorem:

\begin{align*}8^2 + b^2 & = 10^2\\
64 + b^2 & = 100\\
b^2 & = 36\\
b & = \pm 6 \Rightarrow b = 6\end{align*}

(You may have also recognized that this is a “Pythagorean Triple,” 6, 8, 10, instead of using the Pythagorean Theorem.)

You can also find the third side using a trigonometric ratio. Notice that the missing side, \begin{align*}b\end{align*}

\begin{align*}\cos 53.13^\circ & = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{b}{10}\\
0.6 & = \frac{b}{10}\\
b & = 0.6(10) = 6\end{align*}

2. Solve the triangle shown below.

** **

In this triangle, we need to find the lengths of two sides. We can find the length of one side using a trig ratio. Then we can find the length of the third side by using a trig function with the information given originally and a different trig function. Because the side we found is an *approximation*, using the Pythagorean Theorem would not yield the most accurate answer for the other missing side. Therefore, we should use a trig function with the original information to find the length of the third side instead. *Only use the given information when solving right triangles.*

We are given the measure of \begin{align*}\angle A\end{align*}

\begin{align*}\cos 40^\circ & = \frac{adjacent}{hypotenuse} = \frac{6}{c}\\
\cos 40^\circ & = \frac{6}{c}\\
c \cos 40^\circ & = 6\\
c & = \frac{6}{\cos 40^\circ} \approx 7.83\end{align*}

If we want to find the length of the other leg of the triangle, we can use the tangent ratio. This will give us the most accurate answer because we are not using approximations.

\begin{align*}\tan 40^\circ & = \frac{opposite}{adjacent} = \frac{a}{6}\\
a & = 6 \tan 40^\circ \approx 5.03\end{align*}

3. Solve the triangle shown below.

** **

In this triangle, we have the length of one side and one angle. Therefore, we need to find the length of the other two sides. We can start with a trig function:

\begin{align*}\tan 30^\circ & = \frac{opposite}{adjacent} = \frac{b}{7}\\
\tan 30^\circ & = \frac{b}{7}\\
7 \tan 30^\circ & = b\\
b & = 7 \tan 30^\circ \approx 4.04\end{align*}

We can then use another trig relationship to find the length of the hypotenuse:

\begin{align*}\sin 30^\circ & = \frac{opposite}{hypotenuse} = \frac{4.04}{c}\\
\sin 30^\circ & = \frac{4.04}{c}\\
c \sin 30^\circ & = 4.04\\
c & = \frac{4.04}{\sin 30^\circ} \approx 8.08\end{align*}

### Examples

#### Example 1

Earlier, you were asked how far the ball is from the pole.

From our knowledge of how to solve right triangles, we can set up a triangle with the rope and the pole, like this:

From this, it is straightforward to set up a trig relationship for sine that can help:

\begin{align*}
\sin 35^\circ = \frac{opposite}{1}\\
(1)\sin 35^\circ = opposite\\
opposite \approx .5736
\end{align*}

#### Example 2

Solve the triangle shown below:

Since the angle given is \begin{align*}40^\circ\end{align*}

\begin{align*}
\tan 40^\circ = \frac{9}{a}\\
a = \frac{9}{\tan 40^\circ}\\
a = \frac{9}{.839}\\
a = 10.73\\
\end{align*}

We can then use another trig function to find the length of the hypotenuse:

\begin{align*}
\sin 40^\circ = \frac{9}{c}\\
c = \frac{9}{\sin 40^\circ}\\
c = \frac{9}{.643}\\
c = 13.997\\
\end{align*}

Finally, the other angle in the triangle can be found either by a trigonometric relationship, or by recognizing that the sum of the internal angles of the triangle have to equal \begin{align*}180^\circ\end{align*}

\begin{align*}
90^\circ + 40^\circ + \theta = 180^\circ\\
\theta = 180^\circ - 90^\circ - 40^\circ\\
\theta = 50^\circ\\
\end{align*}

#### Example 3

Solve the triangle shown below:

** **

Since this triangle has two sides given, we can start with the Pythagorean Theorem to find the length of the third side:

\begin{align*}
a^2 + b^2 = c^2\\
8^2 + b^2 = 17^2\\
b^2 = 17^2 - 8^2\\
b^2 = 289- 64 = 225\\
b = 15\\
\end{align*}

With this knowledge, we can work to find the other two angles:

\begin{align*}
\tan \angle{B} = \frac{15}{8}\\
\tan \angle{B} = 1.875\\
\angle{B} = \tan^{-1} 1.875 \approx 61.93^\circ\\
\end{align*}

And the final angle is: \begin{align*}
180^\circ - 90^\circ - 61.93^\circ = 28.07^\circ
\end{align*}

#### Example 4

Solve the triangle shown below:

** **

There are a number of things known about this triangle. Since we know all of the internal angles, there are a few different ways to solve for the unknown sides. Here let's use the \begin{align*}60^\circ\end{align*}

\begin{align*}
\tan 60^\circ = \frac{a}{4}\\
a = 4 \tan 60^\circ\\
a = (4)(1.73) = 6.92\\
\end{align*}

and

\begin{align*}
\cos 60^\circ = \frac{4}{h}\\
h = \frac{4}{\cos 60^\circ}\\
h = \frac{4}{.5}\\
h = 8\\
\end{align*}

So we have found that the lengths of the sides are 4, 8, and 6.92.

### Review

Use the picture below for questions 1-3.

- Find \begin{align*}m\angle A\end{align*}
m∠A . - Find \begin{align*}m\angle B\end{align*}
m∠B . - Find the length of AC.

Use the picture below for questions 4-6.

- Find \begin{align*}m\angle A\end{align*}
m∠A . - Find \begin{align*}m\angle C\end{align*}
m∠C . - Find the length of AC.

Use the picture below for questions 7-9.

- Find \begin{align*}m\angle A\end{align*}
m∠A . - Find \begin{align*}m\angle B\end{align*}
m∠B . - Find the length of BC.

Use the picture below for questions 10-12.

- Find \begin{align*}m\angle A\end{align*}
m∠A . - Find \begin{align*}m\angle B\end{align*}
m∠B . - Find the length of AB.

Use the picture below for questions 13-15.

- Find \begin{align*}m\angle A\end{align*}
m∠A . - Find \begin{align*}m\angle C\end{align*}
m∠C . - Find the length of BC.
- Explain when to use a trigonometric ratio to find missing information about a triangle and when to use the Pythagorean Theorem.
- Is it possible to have a triangle that you must use cosecant, secant, or cotangent to solve?
- What is the minimum information you need about a triangle in order to solve it?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 1.10.

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cosine

The cosine of an angle in a right triangle is a value found by dividing the length of the side adjacent the given angle by the length of the hypotenuse.sine

The sine of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the hypotenuse.Tangent

The tangent of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the side adjacent to the given angle.### Image Attributions

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