Lesson 9: the Sine Function

Learning Math With

TerRapin Logo:

Lessons 9-15

By Kathryn Shafer

Updated fall 2017

Table of Contents

Lesson 9: The Sine Function

Lesson 10: Obtuse Scalene Triangles

Lesson 11: Inverse Sine Function

Lesson 12 Acute Scalene Triangles

Lesson 13: Parallelograms

Lesson 14: Kites

Lesson 15: Transformations

Lesson 9: The Sine Function

Purpose: This lesson introduces the sine function, which is used to find an unknown side length in a given right triangle.

Essential Question: Given a right triangle, three known angle measurements and one known side length, how can the sine function and the Pythagorean Theorem be used to calculate the missing side lengths?

A calculator can be used to find the sine of different angle measurements. Scientific calculators have more than one mode setting; so make sure that the mode is set to degrees. Some calculators require you to enter the sine button and thenthe angle’s degrees while others need the degree measurement entered first. Practice with your calculator on the following problems before completing Example 1.

sin(20°) ≈ .342 / sin(45°) ≈ .707 / sin(60°) ≈ .866 / sin(30°) = .500

It is important to round results to at least three decimal places when calculating trigonometric functions that need rounding, because the values are so close together. If the sines of 45°, 46°, 47°, and 48° are rounded to the tenths place they would all have the same value of 0.7. There are two additional trigonometric functions called cosine and tangent. They are commonly found in most high school geometry textbooks and are omitted in this series of lessons.

Example 1: Triangle PQR has angle measurements 30°, 60°, 90°. The short side measures 20 units. Find the measurements of the other two sides.

The sine function is a trigonometric function.
The sine of angle P equals the ratio of RQ to PQ.
sin(angle) = opposite side length / hypotenuse length /
Figure 9.1
Note that sin(30°) = .5
Computation to find h:
sin(30°) = 20/h
.5 = 20/h
.5 * h = 20
h = 40 units

Since we calculated the hypotenuse of triangle PQR as 40 units, we can now use the Pythagorean Theorem to find the third side length.

Computation to find m(PR): / Logo Code / Completed graphic
402 = 202 + PR2
1600 = 400 + PR2
1200 = PR2
SQRT(1200)= PR
34.6 units ≈ PR / fd 34.6
rt (180-30)
fd 40
rt (180-60)
fd 20 /
Figure 9.2

LESSON 9 – TASK 1: Find the missing side lengths for the given triangles.

The solutions are on the next page.

LESSON 9 – TASK 2: Find the measures of the remaining side and angle for each of the triangles in TASK 1.

LESSON 9 – TASK 3: Write the Logo code for each of the triangles in Task 1. Test the code on your computer.

A. Figure 9.3 - Start at Pont L with setheading 0 and left turns

B. Figure 9.4 - Start at point B with setheading 55 and right turns

C. Figure 9.5 - Start at point G withsetheading300 and right turns

D. Figure 9.6 - Start at point K with setheading -25 and left turns

/ Lesson9 Project
Sketch a right triangle on your paper and measure one acute angle, the hypotenuse and one leg. Calculate the measure of the missing side lengths and angles. Write and test the Logo code.

Task 1 and Task 2 Solutions

Figure 9.3
sin(54°) = m/17
sin (54°) * 17 = m
m is approximately equal to 13.8 pixels
m ≈ 13.8 pixels
Segment DE is 9.9 pixels long
SQRT (172 – 13.82) = 9.9 pixels / Figure 9.4
sin(35°) = x/20
sin (35°) * 20 = x
x is approximately equal to 11.5 pixels
x ≈ 11.5 pixels
Segment AB is 16.4 pixels long
SQRT (202 – 11.52) = 16.4 pixels
Figure 9.5
sin(36°) = 10/t
sin (36°) * t = 10
t = 10/[sin(36)]
t is approximately equal to 17 pixels
t ≈ 17 pixels
Segment FG is 13.7 pixels
SQRT (172 – 102) = 13.7 pixels / Figure 9.6
sin(28°) = 50/w
sin (28°) * w = 50
w = 50/[sin(28)]
w is approximately equal to 106.5 pixels
w ≈ 106.5 pixels
Segment IJ is 94 pixels
SQRT (106.52 – 502) = 94 pixels

Lesson 10: Obtuse Scalene Triangles

Purpose: You will use the sine function and the Pythagorean Theorem to find missing side measurements in obtuse scalene triangles. Logo will be used to check your work.

Essential Questions: How can an obtuse scalene triangle be decomposed into two right triangles? How can the sine function and the Pythagorean Theorem be used to create the Logo code for an obtuse scalene triangle?

Please review the material in Lesson 7 before completing this lesson.

Example 1: Obtuse Scalene Triangle XYZ

In this example, you will use both the sine function and the Pythagorean Theorem (PT) to create the Logo code for an obtuse scalene triangle. Triangle XYZ has angles that measure 116°, 27° and 37°. This is clearly an obtuse scalene triangle. In the diagram below, an auxiliary line was drawn perpendicular from X to segment ZY specifically to form two right triangles (∆XPZ and ∆XPY). The purpose of drawing the auxiliary line was to decompose the obtuse scalene triangle into two right triangles. A segment drawn from a vertex perpendicular to the opposite side of a triangle is called the altitude.

In this example, I selected the three angle measures and I made segment XP = 35 units. You might wonder why I had to pick a measure for only one of the sides. If you think about it, the angle measures will determine the general shape of the triangle, but here are an infinite number of similar triangles (of different sizes) that will have angles that measure 116°, 27° and 37° degrees. The length of 35 for the altitude will determine the remaining dimensions.

Figure 10.1

Notice that all of the angle measurements are given for triangle XYZ. Since segment XP is perpendicular to segment ZY, we know that ∠XPZ and ∠XPY are right angles. Thus m(∠PXZ) is 53° and m(∠PXY) is 63° (the sum of the measures of angles in a triangle is 180°). The calculations used to find the missing side lengths and the Logo code are on the next page (start at point Z).

Computation for triangle XPZ / Computation for triangle XPY
sin(37°) = 35/a
.602 ≈ 35/a
.602 * a ≈ 35
a ≈ 35/.602
a ≈ 58.1 units
b2 = 58.12 – 352
b2 = 2150.61
b ≈ 46.4 units / sin (27°) = 35/c
.454 ≈ 35/c
.454 * c ≈ 35
c ≈ 35/.454
c ≈ 77.1 units
d2 = 77.12 – 352
d2 = 4719.41
d ≈ 68.7 units

Completed Figure

Figure 10.2

Finish the problem by creating the logo code and graphic.

Logo Code Template / Logo Code / Completed Graphic
RT (90 – 37)
FD 58.1
RT (180 – 116)
FD 77.1
RT (180 – 27)
FD (68.7 + 46.4) / RT 53
FD 58.1
RT 64
FD 77.1
RT 153
FD 115.1 /
Figure 10.3

Example 2: Obtuse Scalene Triangle QRS

In this example, I will walk through the stepsused to complete calculations for triangle QRS. Let m(QT) = w. Let m(RT) = x. Let m(RS) = y

Figure 10.4

My decision-making process generally depends on the following guidelines:

A. Use the theorem that says the sum of the measures of a triangle equals 180°.

B. If I am given one side length of a triangle, I use the sine function with the measure of the angle opposite the given side length.

C. If I have two side lengths of the triangle, I use the PT with the two given lengths.

Step One – I see there are two right triangles (because of the auxiliary line drawn from point P perpendicular to segment QS). Apply guideline A.

m(∠RQT) = 40° and m(∠SRT) = 65°

Step Two – I see the hypotenuse of triangle RTQ measures 120 pixels. Apply guideline B.

sin (50°) = w/120sin (50°) * 120 = w92 pixels ≈ w

Step Three – I see two of the three side lengths for triangle RTQ. Apply guideline C.

SQRT(1202 – 922)= x77 pixels ≈ x

Step Four – I see the leg of triangle RTS measures 77 pixels. Apply guideline B.

sin (25°) = 77/ysin (25°) * y = 77182 ≈ y

Step Five – I see two of the three side lengths for triangle RTS. Apply guideline C.

SQRT(1822 – 772)= z165 pixels ≈ z

Logo Code Template / Logo Code / Completed Graphic
RT 50
FD 120
RT (180-115)
FD 182
RT (180-25)
FD (165 + 92) / RT 50
FD 120
RT 65
FD 182
RT 155
FD 257 /

LESSON 10 - TASK 1: Write and test the Logo code for the obtuse triangle show here.

Figure 10.6

LESSON 10 - TASK 2: Write and test the Logo Code for the obtuse triangle shown here.

Figure 10.7
/ Lesson10 Project
Write and test the Logo code for the shape shown in figure 10.8. Use the grid to approximate segment lengths.
Hint – there are matching obtuse isosceles triangles.

Figure 10.8

Lesson 11: Inverse Sine Function

Purpose: This lesson introduces the inverse sine function which is used to find an angle measure in a given right triangle.

Essential Question: Given a right triangle and three known side measurements, how can the inverse sine function be used to calculate the missing angle measures?

What would happen if all of the side measurements of a right triangle are given, but the two acute angle measures are unknown? How would you find the two missing angle measures?

Let’s revisit the mathematics in a right triangle. Using what you know about Pythagorean triples, let the side lengths of triangle PAQ be 30, 40 and 50 turtle steps. Since these lengths are the Pythagorean triple 3, 4, 5, times a factor of 10, these side measurements will form a right triangle.

Greek letters are often used in mathematics as a variable for an unknown angle measurement. In the figure below the letters alpha (α) and beta (β) are used to indicate measures of the exterior angles at points P and Q, respectively.

The goal is to write the Logo code (starting at point A in a clockwise direction).

Figure 11.1 / Logo Code Template
FD 30
RT ? (exterior angle at point P)
FD 50
RT ? (exterior angle at point Q)
FD 40
Try to estimate the angle measures before reading the next section.

Note that you do not know either the interior or the exterior angles at points P and Q. But, since this is a right triangle you can use the sine function to find the triangle’s interior angles. Recall that the sine function uses three values, one of the non-right angles, the opposite side length and the hypotenuse. The set up looks like this:

sin(m(∠P)) = 40/50
In words, this reads “the sine of the measure of angle P equals 40 divided by 50.”
To “undo” the sine function I must take the inverse.
The inverse symbol for the sine function is written as “sine to the negative one power” or sin-1.
The next step looks like this:

sin-1 [sin(m(∠P))] = sin-1(40/50)
When I take the inverse of the sine function on the left side of this equation, I must do so on the other side of the equation!

Simplify the equation: sin-1 [sin(m(∠P))] = sin-1(40/50)

m(∠P) = sin-1 (40/50)
m(∠P) ≈ 53.1° (tenths is sufficient)
Repeat these steps to find the measure of angle Q.

sin(m(∠Q)) = 30/50
sin-1 (sin(m(∠Q))) = sin-1 (30/50)
m(∠Q) = sin-1 (30/50) ≈ 36.9°

A good check of the calculations would be to add 36.9° and 53.1°, to confirm that the total is 90°. You may have noticed that subtracting 53.1° from 90° could have been done to find the measurement of angle Q, because the sum of the angle measures for any triangle has to be 180°.

Return to the diagram and write in the angle measurements. To complete the Logo code, find the angle of turn for the turtle as you trace the path from point A clockwise around the triangle.

Figure 11.2 / Logo Code Template
Let the variables alpha and beta represent the missing angle measurements.
FD 30
RT α
FD 50
RT β
FD 40
Solution
Find α
(180 – 53.1) = 126.9°
Find β
(180 – 36.9) = 143.1° / Completed Logo Code:
FD 30
RT 126.9
FD 50
RT 143.1
FD 40 /
Figure 11.3

LESSON 11 - TASK 1:

Following the process described above, find the missing angle measures and write the Logo code for a right triangle with side measurements 30 cm, 59.9 cm, and 67 cm.
A sketch of this triangle is shown to the right.
I suggest that you start at point E with a startingheading of 350°(move clockwise). /
Figure 11.4

Obtuse Scalene Triangle EFG [see figure 11.5 below]

In your job as a learner or teacher, you might have to approximate measurements for a triangle that already exists on a worksheet or textbook page. Triangle EFG appears to be an obtuse scalene triangle. Drop a perpendicular segment from point E to segment GF and apply what you know about right triangles. The measure of ∠EPF is 90° because segment EP is perpendicular to segment GF.

Look at triangle EFP and estimate that segment EP is longer than segment PF.

Let EP = 25 units, and let PF = 20 units. Be careful to only pick two sides in this triangle!

Figure 11.5

Calculate the missing measurements in triangle EPF.

252 +202 = EF2
625 + 400 = EF2
1025 = EF2
32.0 units ≈ EF / m(∠F) = sin-1 (25/32)
m(∠F) ≈ 51.4° / m(∠FEP) = 90 – 51.4
m(∠FEP) = 38.6°

Figure 11.6

Now find the missing measurements in triangle EPG. We may only pick one measurement. Let GP = 60 units since it appears to be about three times as long as segment PF. (Note that we could have picked one of the missing angle measurements.)

602 + 252 = EG2
3600 + 625 = EG2
4225 = EG2
65 units ≈ EG / m(∠G) = sin-1 (25/65)
m(∠G) ≈ 22.6° / m(∠GEP) = 90 – 22.6
m(∠GEP) = 67.4°

A quick check indicates that m(∠GEF) totals 106°, forming an obtuse angle. If m(∠GEF) was not greater than 90°, the side measurement of 60 units would have to be increased and the last three calculations repeated.

Figure 11.7 (in this figure, angle G was rounded up from 22.6 degrees)

To finish this example, the Logo code has been written as follows. Since the turtle will have to turn before drawing side GE, the code begins with a right turn at point G. In figure 11.7, the measures of ∠G and ∠F were rounded to 23° and 51° respectively. The Logo code follows the calculations shown above.

Logo Code Template / Logo Code / Completed Graphic
RT (90 – 22.6)
FD 65
RT (180 – 106)
FD 32
RT (180 – 51.4)
FD (20 + 60) / RT 67.4
FD 65
RT 74
FD 32
RT 128.6
FD 80 /
Figure 11.8

Common Mistake Number One:

Selecting conflicting measurements!

The most common error pattern noticed when working with these types of problems is the learner who picks too many measurements. This will cause a conflict within the shape!

CAUTION: While rounding is not an error, the cumulative effect of repeated rounding will cause the measurements to be off by enough that the turtle will not end up back at the starting point. Therefore, it is a good idea to round to the tenths place for both the side and angle measurements.

LESSON 11 - TASK 2: Complete the Logo code for the obtuse triangle show here.

Figure 11.9
/ Lesson 11 Project
Choose the appropriate number of side lengths in the graphic shown below. Use the inverse sine function to find the missing angles. Write and test the logo code.

Figure 11.10

Lesson 12 Acute Scalene Triangles

Purpose: This lesson focuses in decomposing acute scalene triangles into right triangles. You will use what you know about the Pythagorean Theorem. The 45-45-90 and the 30-60-90 triangles are introduced in this lesson. Knowledge of the sine function is necessary in the problems marked with an asterisk.

Essential Question: Can you apply what you have learned about the sum of the measures of a triangle and the Pythagorean Theorem to solve complex problems?

*Example 1 – Acute Scalene Triangle JKM

In this example, you will use both the sine function and the Pythagorean Theorem to write the Logo code for an acute scalene triangle (example 2 does not require the sine function). Begin the Logo code at Point J. Triangle JKM has three acute angles and an altitude has been drawn from point K to segment JM. The calculations have been rounded to the hundredths place to avoid rounding error.

Figure 12.1
Work for Triangle KTJ, on Left / Work for Triangle KTM, on Right
y2 = 2002 + 2002
y2= 40,000 + 40,000
y ≈ SQRT(80,000) ≈ 282.84 pixels
Note that y = 200 * SQRT (2) = 282.84
The leg measure times the SQRT (2) equals the measure of the hypotenuse. This is true for any 45-45-90 triangle. / sin(60) = 200 / z
z = 200 / sin(60°)
z ≈ 230.94 pixels
230.942 = 2002 + w2
230.942 - 2002 = w2
115.47 units ≈ w

There is a special relationship between the three side lengths of triangle KTM. See if you can find it. The solution is at the end of this lesson.

Logo Template / Logo Drawing
RT 45
FD 282.84
RT (180 - 75)
FD 230.94
RT (180 - 60)
FD (115.47 + 200) /
Figure 12.2

Example 2: Acute Scalene Triangle ABC

An altitude from point C to side AB in triangle ABC is given in figure 12.3. This segment decomposes triangle ABC into two right triangles. Compute the missing angle measures and side lengths, then write/test the Logo code. The work is on the next page.

Figure 12.3
x2+4.32 = 8.22
x2= 8.22 – 4.32
x2= 67.24 – 18.49
x = SQRT(48.75)
x≈7.0 cm
w2 = 9.62 + 72
w= SQRT( 92.16 + 49)
w ≈ 11.9 cm
α = 180° – (85.1° + 58.6°)
α = 36.3° /
Figure 12.4

To finish the problem, write the logo code starting at point B, moving up and turning counterclockwise or left. Starting at point B is easier than trying to find the initial turn at point C. A scale factor of 10 is used for each of the side lengths, because the side lengths are too small to use in the Logo code.

Calculation Template / Simplified Logo Code / Logo Drawing
FD 96 + 43
LT (180 – 36.3)
FD 119
LT (180 – 85.1)
FD 82 / FD 139
LT 143.7
FD 119
LT 94.9
FD 82 /
Figure 12.5

LESSON 12 - TASK 1: Calculate the three missing side lengths for acute scalene triangle ABC. Write and test the Logo code.

Figure 12.6
/ LessonTwelve Project
Sketch an acute scalene triangle on your paper. (This can be tricky)
Measure the side lengths with a ruler and the angle measures with a protractor. Use a scale factor of 10 to write/test the Logo code.
You may NOT use the same measurements shown in the examples above.

Discussion of Example #1 – The 30 – 60 – 90 triangle

The measure of the longer leg (across from the 60 degree angle) is equal to the short leg times the square root of 3.In the example, we see that (115.47 * SQRT (3)) ≈ 199.99units or 200 units when rounded.

The hypotenuse is equal to the short leg times 2. In the example, we see that 115.47 * 2 = 230.94 units.

Lesson 13: Parallelograms

Purpose: This lesson introduces the definitions and properties of parallelograms. The Logo program provides an excellent environment to explore different types of parallelograms. This lesson uses a procedure with three variables to efficiently draw a given parallelogram.

Essential Question: Can you apply your knowledge of interior and exterior angle measures to code and investigate the properties of parallelograms?

Definitions:

A quadrilateral is a closed polygon with exactly four sides.

A parallelogram is a quadrilateral that has two pairs of parallel sides.

A rectangle is a parallelogram with a right angle.

A square is a rectangle with all sides of equal length (equilateral).

A rhombus is a parallelogram with all sides of equal length (equilateral).

Recall that a regular polygon is both equiangular and equilateral. The square (introduced in Lesson #5) is the only regular quadrilateral.

Note that the precise wording of a definition will vary depending on the book you are using. For example, a square could be defined as a rhombus with a right angle.

LESSON 13 – TASK 1:

The graphic in Figure 13.1 shows parallelogram ABCD with given side lengths of 135 pixels by 195 pixels and an acute interior angle of 40 degrees. An auxiliary line was drawn from point A perpendicular to line BC. Write and test the Logo code for this shape starting at point A. (check your work at the end of this lesson).

Figure 13.1

A parallelogram provides a good example of pairs of lines that are cut by a transversal. For example, lines AB and DC are cut by transversal AD (and transversal BC). Lines AD and BC are cut by transversal AB (and transversal DC). It helps to use colored pencils when locating a transversal line!