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1.8: Right Triangle Trigonometry

Difficulty Level: At Grade Created by: CK-12
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Day 1 Day 2 Day 3 Day 4 Day 5

The Pythagorean Theorem

Investigation 8-1

The Pythagorean Theorem Converse

Quiz 1

Start Similar Right Triangles

Finish Similar Right Triangles

Special Right Triangles

Investigation 8-2

Day 6 Day 7 Day 8 Day 9 Day 10

Finish Special Right Triangles

Investigation 8-3

Quiz 2

Start Tangent, Sine, and Cosine Ratios

Finish Tangent, Sine, and Cosine Ratios Solving Right Triangles Finish Solving Right Triangles
Day 11 Day 12 Day 13

Quiz 3

Start Review of Chapter 8

Review Chapter 8 Chapter 8 Test Start Chapter 9

Be sure to take your time with this chapter. Remember that the Pacing Guide is merely a suggestion. Students can get caught up with vocabulary and theorems in this lesson.

The Pythagorean Theorem


This lesson introduces the Pythagorean Theorem. It has several applications which we will explore throughout this chapter.

Relevant Review

Thoroughly review the Simplifying and Reducing Radicals subsection in this lesson. In this and subsequent lessons, answers will be given in simplest radical form. Make sure students know how to add, subtract, multiply, divide, and reduce radicals before moving on. Present anything under the radical like a variable. Students know they cannot add \begin{align*}2x + 5y\end{align*}. Therefore, they cannot combine \begin{align*}2 \sqrt{3} + 5 \sqrt{2}\end{align*}. This should make it easier for students to understand.

Teaching Strategies

Investigation 8-1 provides one proof of the Pythagorean Theorem. First, lead students through steps 1-3, then have them finish step 4 individually while you circle around to answer questions. Once students are done with this investigation, either take students to the computer lab or if you have an LCD screen, go to the site in the FlexBook to see two additional proofs of the Pythagorean Theorem. Both of these proofs are animated and provide another viewpoint.

There are several applications of the Pythagorean Theorem. Students need to be familiar with as many as possible. Go over each example to make sure students understand the range of questions that can be asked. Each of the examples, even though worded differently, are completed the same way. The directions for a given problem can be one place where students can have confusion. Going over the different types of directions for the same type of problem can be very helpful.

Finally the Distance Formula is proven in this lesson. Students might already have an idea of how it works. Explain that it is a variation on the Pythagorean Theorem, especially the before we solve for \begin{align*}d\end{align*}, when the equation looks like \begin{align*}d^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2\end{align*}. Notice in this section, the Distance Formula looks different than it did when it was introduced in Chapter 3. Recall that order does not matter, as long as the corresponding \begin{align*}x\end{align*} or \begin{align*}y\end{align*} value is first (\begin{align*}x_1\end{align*} and \begin{align*}y_1\end{align*} are first or \begin{align*}x_2\end{align*} and \begin{align*}y_2\end{align*} are first).

Converse of the Pythagorean Theorem


This lesson applies the converse of Pythagorean’s Theorem to determine whether triangles are right, acute, or obtuse.

Relevant Review

Make sure students are comfortable squaring and simplifying radical numbers.

Teaching Strategies

The Converse of the Pythagorean Theorem basically says if the sides of a triangle do not satisfy the Pythagorean Theorem, then the triangle is not a right triangle. Theorems 8-3 and 8-4 extend this concept to determine if the non-right triangle is acute or obtuse. To help students remember which is which, tell them to think opposite. When \begin{align*}a^2 + b^2 > c^2\end{align*} (\begin{align*}a^2 + b^2\end{align*} is greater than \begin{align*}c^2\end{align*}) the triangle is acute (all angles are less than \begin{align*}90^\circ\end{align*}). When \begin{align*}a^2 + b^2 < c^2\end{align*} (\begin{align*}a^2 + b^2\end{align*} is less than \begin{align*}c^2\end{align*}) the triangle is obtuse (one angle is greater than \begin{align*}90^\circ\end{align*}). Also, when in doubt, have them draw the triangle, as best they can, to scale. Then, they can see what the triangle should be and they can do the Pythagorean Theorem to confirm or deny.

Remind students that the Triangle Inequality Theorem still holds. So, if they are given lengths like 5, 7, and 15, they need to be able to recognize that these lengths do not make a triangle at all. \begin{align*}5 + 7 > 15\end{align*}, so no triangle can be formed. If students do not see this, they will be doing unnecessary work, not to mention think that the triangle is obtuse. Review the Triangle Inequality Theorem as you are going through the examples in the text. No one likes to do extra work, if they do not have to.

Additional Example: Do the following lengths form a triangle? If so, is it acute, right, or obtuse?

a) 10, 15, 20

b) 7, 14, 21

c) \begin{align*}8\sqrt{2}, 4 \sqrt{6}, 4 \sqrt{14}\end{align*}

Solution: First, check all the lengths to see if they make a triangle. b) does not, \begin{align*}7 + 14 = 21\end{align*}, so those lengths cannot make a triangle. Let’s see what type of triangles a) and c) are.

a) \begin{align*}&10^2 + 15^2 = 20^2\\ &100 + 225 < 400\\ &\text{obtuse triangle}\end{align*}

c) \begin{align*}&(8\sqrt{2})^2 + (4\sqrt{6})^2 = (4 \sqrt{14})^2\\ &64 \cdot 2 + 16 \cdot 6 = 16 \cdot 14\\ &128 + 96 = 224\\ &\text{right triangle}\end{align*}

Additional Example: Find an integer such that 9, 12, ____ represent an acute or obtuse triangle.

Solution: 9, 12, 15 would be a right triangle (this is a multiple of the Pythagorean triple, 3-4-5). So 8, 9, 10, 11, 12, 13, or 14 would work. If the integer is less than 8, then the triangle would be obtuse, with 12 as the longest side.

For an obtuse triangle, the third side could be less than 8, but greater than \begin{align*}3 (9 + 3 = 12)\end{align*}. And, it could also be greater than 15, but less than \begin{align*}21 (9 + 12 = 21)\end{align*}. So, the possibilities are 4, 5, 6, 7, 16, 17, 18, 19, or 20.

Using Similar Right Triangles


Students will review similar triangles, primarily right triangles. Then, the concept of the geometric mean is introduced and applied to right triangles.

Relevant Review

Review similar triangles and their properties briefly before introducing Theorem 8-5. The two triangles formed by the altitude from the right angle in a right triangle are similar to the larger right triangle by AA (see Example 1). Show students how the three triangles fit together and which angles are congruent to each other and which sides are proportional (Examples 1 and 2).

Also, remind students that answers should be in simplest radical form. You may need to add a few simplifying and reducing radical questions in the Review Queue.

Teaching Strategies

Example 3 introduces the geometric mean as it applies to right triangles. You can choose to use this example before discussing the geometric mean or after it, after Example 7. It might be easier for students to see the geometric in its literal, algebraic form (without being applied to triangles) and practice it that way for a few examples, and then apply it to a right triangle.

In Examples 5 and 6, it might be helpful to show students the corresponding proportions for the geometric mean, \begin{align*}\frac{24}{x} = \frac{x}{36}\end{align*} and \begin{align*}\frac{18}{x} = \frac{x}{54}\end{align*}. Students will like the short cut, \begin{align*}x = \sqrt{ab}\end{align*}, but they should be shown the proportion first. The proportion directly relates to its application to right triangles.

Examples 3 and 7 apply the geometric mean to the right triangle. The set-up of these proportions, using similar triangles, is the same as the geometric mean. So, students can always use similar triangles, rather than memorize the geometric mean.

So often in math, there are two or even three ways to solve one problem. In Example 8, there are two ways presented to solve for \begin{align*}y\end{align*}. Encourage students to try the geometric mean, but they can also use the Pythagorean Theorem. Show students both methods and then brainstorm why one method could be preferred over the other. Students should be allowed to use which ever method they feel more comfortable with.

Special Right Triangles


The purpose of this lesson is to encourage the use of ratios to find values of the sides of special right triangles. These triangles are extremely useful in trigonometry.

Relevant Review

Special right triangle ratios are extended ratios. Here is an additional Review Queue question:

Additional Review Queue: The sides of a triangle are in the extended ratio 3:7:9. If the shortest side is 21, find the length of the other two sides.

Solution: We will rewrite the ratio as \begin{align*}3x: 7x: 9x\end{align*}. So, \begin{align*}3x = 21\end{align*}, which means \begin{align*}x = 7\end{align*}. Therefore, the other two sides are \begin{align*}7 \cdot 7 = 49\end{align*} and \begin{align*}9 \cdot 7 = 63\end{align*}.

Teaching Strategies

Students should complete Investigation 8-2 independently. Lead students through step 1, but then allow them to complete the Pythagorean Theorem in steps 2 and 3 on their own. Ask students if they see a pattern among the hypotenuses. Then, go over how to solve an isosceles triangle given various sides and with various square roots. They should know how to solve an isosceles right triangle when the legs are not whole numbers. To find the legs, always divide the hypotenuse by \begin{align*}\sqrt{2}\end{align*} and then simplify the radical. (Students might also notice there is a pattern here too. The leg will be the hypotenuse divided by 2 and multiplied by \begin{align*}\sqrt{2}\end{align*}. For example, if the hypotenuse is 20, the legs will be \begin{align*}20 \div 2 \cdot \sqrt{2} = 10 \sqrt{2}\end{align*}. This is a little short cut.) To find the hypotenuse, multiply the leg by \begin{align*}\sqrt{2}\end{align*}. Students may need to simplify the square root. Also, the diagonal of a square will always split the square into two 45-45-90 triangles. Make sure students notice this connection.

With 30-60-90 triangles, students must make the connection that it is half an equilateral triangle. In Investigation 8-3, lead students through steps 1-3, then allow the students complete steps 4 and 5 on their own. Students will have a hard time remembering this extended ratio. Some students will think that \begin{align*}x\sqrt{3}\end{align*} should be the hypotenuse because 3 is bigger than 2. However, explain that \begin{align*}\sqrt{3} \approx 1.73\end{align*}, which is smaller than 2, so \begin{align*}2x\end{align*} will always be the longest side, which is the hypotenuse. To solve 30-60-90 triangles, students will need to find the shortest leg, if they are not given it. If they are given the hypotenuse, divide by 2, then they can multiply by \begin{align*}\sqrt{3}\end{align*} to get the longer leg. If they are given the longer leg, they will need to divide by \begin{align*}\sqrt{3}\end{align*} (or use the shortcut as applied to the 45-45-90 triangle above, use 3 and \begin{align*}\sqrt{3}\end{align*} where appropriate) to get the shorter leg, then multiply that by 2.

Students will also get these two ratios confused. One way to help them remember is for a 45-45-90 triangle there are 2 \begin{align*}45^\circ\end{align*} angles, so the hypotenuse is \begin{align*}x\sqrt{2}\end{align*}. For a 30-60-90 triangle all the angles are divisible by 3, so the \begin{align*}\sqrt{3}\end{align*} is in the radical for this ratio.

The best way for students to become comfortable with these ratios is to have them do lots of problems. Students can also make flashcards for these ratios. If students have trouble remembering these special shortcuts, encourage them to use Pythagorean’s Theorem and simplify the answer. The resulting answer will equal the shortcut.

Tangent, Sine and Cosine


This lesson introduces the trigonometric functions; sine, cosine and tangent.

Teaching Strategies

Before introducing the trig ratios, make sure students understand what adjacent and opposite mean and which angles they are in reference to. \begin{align*}c\end{align*} will always be the hypotenuse, but \begin{align*}a\end{align*} and \begin{align*}b\end{align*} can be either opposite or adjacent, depending on which acute angle we are using. At this point, do not overwhelm students with the fact that the trig functions can be applied to any angle; focus on acute angles in triangles.

Encourage students to make flash cards for the sine, cosine and tangent ratios and to use the pneumonic SOH-CAH-TOA (in FlexBook). Both of these things will help students internalize the ratios.

In the types of problems in this lesson, it will be very common that two of the three sides are given and students will need to use the Pythagorean Theorem to find the third side. This should always be done first, and then they can apply the ratios. Students will also need to reduce ratios and simplify any radicals. Show students several different orientations of the triangles (rotated, flipped, etc) so they are familiar with where an angle is and which sides are adjacent and opposite.

Have each student check to ensure their calculator is set to degrees (DEG), not radians (RAD). Having a calculator in radians will provide incorrect answers. When checking homework at the beginning of the class period, check the mode of students’ calculators as well.

When introducing how to find the sides of a right triangle, using the trig ratios, draw the triangle from Example 5 on the board with only one variable, \begin{align*}a\end{align*}. This will isolate \begin{align*}a\end{align*} and students should be able to see that the cosine ratio will solve for \begin{align*}a\end{align*}. Use the arrow to help illustrate that \begin{align*}a\end{align*} is adjacent to \begin{align*}22^\circ\end{align*} and 30 is the hypotenuse. After \begin{align*}a\end{align*} is found, redraw the triangle so \begin{align*}b\end{align*} is the only variable. Now, \begin{align*}b\end{align*} is isolated and students will be able to recognize that the sine ratio will solve for it. Again, use arrows, if needed.

Stress to students that they should only use information that they are given in the problem. Using “solved for” information will not give them the most accurate answer or it could be completely wrong (if the “solved for” answer used is incorrect).

To help illustrate the angles of elevation and depression, see the picture below.

Show this to students and ask what the angle of elevation from the cow to the goat is. Then, ask what the angle of depression from the sheep down to the cow is. Students should notice it is the same measurement and alternate interior angles. Fill in the angle of depression/elevations with any measurement.

Inverse Trigonometric Ratios


In the previous lesson, students used the special trigonometric values to determine approximate angle measurements. This lesson enables students to “cancel” a trigonometric function by applying its inverse to accurately find an angle measurement.

Relevant Review

Begin by listing several mathematical operations on the board in one column. In a second column, title it “Inverse.” Be sure students understand what an inverse means (an inverse cancels an operation, leaving the original value undisturbed).

Operation Inverse Example

The first four are typically easy for students (Subtraction, square root, multiplication, and addition). You may have to lead students a little more on the last two (inverse tangent and inverse sine). Students may say, “Un-tangent it.” Use the correct terminology here, but also use their wording, if at all possible. Students will be able to cancel the trigonometric function using the inverse of that function, even though they may use incorrect terminology.

Go over Example 1 thoroughly. Make sure every student understands how to input inverse trig functions into their calculator. Remind students that the set-up for inverse problems is the same as those from the previous lesson. However, instead of being given an angle measure, we leave it as a variable. Students need to solve for the angle and then put everything into the calculator at the same time. Get them in this habit so they will produce the most accurate answers. For example:

\begin{align*}\underline{\text{Yes}}: \sin^{-1} \left (\frac{2}{3}\right ) && \underline{\text{No}}: \sin^{-1} (0.666)\end{align*}

Example 5 is a special right triangle. Students will probably go through the motions and not notice that they can use the ratios learned in the Special Right Triangles lesson. Either way, students will still arrive at the correct answer, but point out to them to not go blindly into each question. Read a problem, re-read, and then decide how to answer.

Like the last lesson, real-life situations are a major application. At the end of this lesson, create a word problem as a class. Use the names of students or find the height of a local building. Make the problem personal. Then, add it to the test as an extra question or a bonus.

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