Whatcom Math Project
Exploring the Tangent Ratio
Course: Geometry / Algebra 2
College Readiness Standards:
Standard 1: Reasoning/Problem-solving – The student uses
logical reasoning and mathematical knowledge to define
and solve problems.
Standard 3: Connections – The student extends
mathematical thinking across mathematical content
areas, and to other disciplines and real life situations.
Standard 5: Geometry – The student makes hypotheses,
models situations, draws conclusions, and supports claims using geometric
concepts and procedures.
Standard 8: Functions – The student accurately describes and applies function
concepts and procedures to understand mathematical relationships.
Student Attributes:
1 – Demonstrates intellectual engagement.
Student Learning Outcome Components:
1.1 – Analyze a situation and describe the problems to be solved
3.4 – Abstract mathematical models from word problems, geometric problems,
and applications.
5.4 – Recognize and apply the basic right-triangle trigonometric relationships to
solve problems.
8.6 – Recognize, analyze, interpret, and model with trigonometric functions.
Learning Objectives:
5.4.a – Use sine, cosine or tangent to find unknown distances and angles.
5.4.b – Use the inverse of sine, cosine or tangent to find the measure of a
missing angle.
8.6.d – Sketch graphs of tangent function, without technology; identify the
domain, range, intercepts, and asymptotes.
Prerequisite Skills:
Students will need a basic understanding of the basic trigonometric ratios, how to
solve for angles in triangles using the triangle-sum theorem, how to use the
Pythagorean theorem, and how to measure angles with protractor.
Estimated Time for Completion:
4 to 5 days for the entire lesson, but easily broken up into smaller segments.
1
Material for Students:
Protractors, rulers (metric), scientific calculators (with trigonometric functions),
worksheets
Teaching Aids:
Laser pen (optional), calculator.
References:
Eli Maor, Trigonometric Delights
2
I.
PREASSESSMENT:
1) Solve for x:
12
33°
x
a) 7.79 (used adjacent over opposite)
b) 14.31 (used cosine instead of tangent)
c) 18.48 (correct answer)
d) 22.03 (used sine instead of tangent)
2) Solve for the measure of A:
B
18
A
14
C
a) 1.3 (did not take inverse tangent of opposite/adjacent ratio)
b) 22.8 (found the length of AB)
c) 37.9 (used cotangent ratio)
d) 52.1 (correct answer)
3
3) Solve for the length of QR:
Q
65°
R
P
10
a) 5.73 (used cosine)
b) 7.00 (used cotangent)
c) 9.06 (correct answer)
d) 21.44 (used tangent)
4) Calculate the tangent of B.
A
10
6
C
a)
b)
c)
d)
8
B
0.60 (student found sine of B)
0.75 (correct answer)
0.80 (student found cosine of B)
1.33 (student found cotangent of B)
4
5) A ray starts at the origin and passes through the point (12, 5). Solve for the angle
between the x – axis and the ray.
(12, 5)
(0, 0)
a)
b)
c)
d)
67.38
65.38
24.62
22.62
(used inverse cotangent)
(used inverse cosine of 5/12)
(used inverse sine of 5/12)
(correct answer)
5
II.
INTRODUCTION:
This lesson is intended to help students explore the range and behavior of trigonometric
functions, particularly tangent. Geometry is taught after algebra, so student have some
knowledge of non-linear functions, but given the connection between right – triangle
trigonometry and similar triangles, in which proportions are so prevalent, the idea that
the functions are not linear (and indeed have an asymptote for tangent) is not usually
fleshed out. This lesson has students explore the tangent ratio as a function of its angle
from a variety of different perspectives and hands – on activities. It also helps students
use the tangent function in a variety of problem – solving situations, including solving for
missing angles and sides, often using multi-step methods.
6
III.
LESSON:
The main idea for this lesson is the non-linearity of the relationship between the
angle and the tangent. The main lesson divided into 6 parts (Part 0 through 5), which
start with a review of basic trig ratios, and then review how to use the tangent ratio to
solve for unknown sides and angles. The basic method is to have students actually see
the lines and angles and try to visualize tangent as a height of a wall; if we keep the
adjacent side constant (the floor in the scenario), then changing the angle changes the
height. Part 4 then draws these ideas together and has students graph the relationship
between the angles and the heights of the triangles they have drawn. Part 5 prepares
students for Algebra 2 by extending the idea of angles in the negative direction. The
application sections (Parts 6 and 7) are much more teacher-directed, showing how to
use the tangent ratio to solve problems that involve several steps, how to break figures
down into sub-problems, and redraw parts of figures to provide clarity to complicated
diagrams.
The lesson as a whole will probably take 4 to 5 days of a 60-minute class period,
although the parts can be broken up as necessary, depending on how the students are
progressing. The lesson is presented as an integrated whole, with the teacher notes
and student worksheets embedded within the notes, followed by the worksheets
answered and with appropriate graphs or drawings. The worksheets and the
assessments are also collected together at the end of the document for easy
printing.
7
Part 0 – Teacher-led review of trig ratios for right triangle.
hypotenuse
5
opposite

3

adjacent
4
In the right triangles above, the tangent of  is the ratio of the side opposite (across
opposite side
from) the angle, divided by the side adjacent (next) to the angle: tan  
.
adjacent side
3
In the triangle to the left, we write tangent , abbreviated tan , as tan   . This ratio
4
will be constant no matter how large the sides of the triangle are. This means that we

can use the tangent ratio to determine the lengths of unknown sides, as in the figure
below:

B
x
50°
A
9
Here, tan 50  =
C
x
, or, rewriting,
9
9(tan 50)  x
9(1.1917) = x
x = 10.73.


8
We can also use the tangent ratio to calculate unknown angles if we know the opposite
and adjacent sides:
5
3

4
3
. We can take the inverse tangent (usually
4
3 
written tan-1 or arctan) on a calculator (using the “shift” or “inv” key): tan-1   36.87 .
4 
Check and see that students
 can find the missing measures in these examples; be sure
they can set up the proper ratio (opposite / adjacent) and can use the inverse tangent
sequence on their calculators. Also, get students into the habit of checking tan 45 on

their calculator; if they get 1.620, their calculator is set in radians
and will have to be
switched to degree mode.
Two more examples:
In the triangle above, we know that tan  
42°
x
5

18
12
tan  =

5
, so  = 22.69
12
tan 42 =
18
, so x tan 42 =18 , and x = 20.00
x
Note that on this second example, we could also find the measure of the other angle

x

(90 – 42 = 48) and use that angle;
we then get tan 48 =
or x =18 tan 48 ,
18
which is often an easier calculation for students to work with.


9
Part 1 – Working with the tangent ratio
a) We will start the student activity by having them draw a right triangle with any angle
between 10 and 30 degrees, measuring the length of the side opposite the chosen
angle. (Note: the angle should not be greater than 30 degrees, because the second
triangle will be much too large, also the angle should not be smaller than 10 degrees,
because the measurements will not show much more than a doubling of the opposite
side length.)
Ask students: what will happen to the height if you double the angle?
Most students will assume the linearity of the function, that is, if you double the
angle, you will double the height. The main object of the lesson is to elicit this
response and point towards the non-linearity of the tangent function.
b) Then, on the same triangle, with the same adjacent side and same right angle, draw
another triangle with the chosen angle doubled. Students should measure the length of
the side opposite their new angle, and compare it to their original height and their
conjecture.
Ask students: what conclusions can we draw from the activity? What would
happen to the original height if we had tripled the angle? Quadrupled?
Work should look like this:
If we double the angle, we more
than double the length of the
opposite side. The tangent
function is not linear, so if we
triple or quadruple the angle, it
will grow by a factor of more than
3 or 4, respectively.
tan 2
tan



10
Part 2 – Using tangent ratio to compute heights
Materials: Protractors, rulers (metric), calculators, worksheet.
For the second activity, students will be modeling a laser pointed at a wall at various
angles, and measuring the height of the image on the wall (for this purpose, the image
starts at the floor and ends at the intersection of the laser and the wall). If you do not
have a laser pen, a flashlight would work as well to demonstrate the activity. It is not
necessary to have lights for the students.
Directions:
Start with 10 cm distance from laser to wall, as shown on the worksheet.
a) Draw a line with the vertex at the laser, that makes an angle of 10 with the
horizontal line (the floor), continuing until it intersects the wall.
b) Measure the height of the intersection from the base of the wall, and enter the
measurement on the table.
c) Repeat steps a and b for the angles 20, 30, 40, 50, 60, and 70, and
measure the heights from the base to the intersection. Enter your data on the table.
d) Calculate the differences between the heights you have found.
For the teacher:
The last 4 entries will not fit on the paper, and will have to be calculated. It will
be useful to show the following equation on the board as an example:
x
tan 10 
.
10 cm
The students should solve for x:
height 10 (tan10)
or more generally


height 10 (tan )
When you have finished your table:

1) Describe the behavior of the height as the angle approaches 90.
2) Does the height increase at a steady rate?
3) What will happen to the intersection of the laser and the wall when the angle of the
laser is 90?
4) What happens to the differences in height as the angle of the laser gets larger?
11
Directions:
Start with 10 cm distance from
laser to wall, as shown on the
worksheet.
a) Draw a line with the vertex at
the laser that makes an angle of
10 with the horizontal line with
respect to the floor, continuing
until it intersects the wall.
b) Measure the height of the
intersection from the base of the
wall, and enter the measurement
in the table.
c) Repeat steps a) and b) for the
angles 20, 30, 40, 50, 60,
and 70, and measure the
heights from the base to the
intersection. Enter your data in
the table.
d) Calculate the differences
between the heights you have
found.
wall
laser
10 cm
12
Worksheet 1
Angle
Height of image
(in cm)
0
0
Difference in height
(subtract previous
height)
------
10
20
30
40
50
60
70
80
85
89
When you have finished your table:
1) Describe the behavior of the height as the angle approaches 90.
2) Does the height increase at a steady rate?
3) What will happen to the intersection of the laser and the wall when the angle of the
laser is 90?
4) What happens to the differences in height as the angle of the laser gets larger?
13
laser
10 cm
(The lines should look like this.)
14
Worksheet 1---answers
Angle
Height of image
0
0
Difference in height
(subtract previous
height)
------
10
1.76 cm
1.76
20
3.64
1.88
30
5.77
2.13
40
8.39
2.62
50
11.92
3.53
60
17.32
5.40
70
27.47
10.15
80
56.71
29.24
85
114.30
57.59
89
572.6
458.60
When you have finished your table:
1) Describe the behavior of the height as the angle approaches 90.
As the angle approaches 90, the height gets increasingly large.
2) Does the height increase at a steady rate?
No, the rate of increase also increases.
3) What happens to the intersection of the laser and the wall when the angle of the
laser is 90?
At 90 the laser no longer hits the wall.
4) What happens to the differences in height as the angle of the laser gets larger?
The differences in height also get bigger at a faster and faster rate.
15
Part 3 – Inverse Tangents
Worksheet 2
Teacher notes:
Have students measure a height on the wall of 15 cm, then draw the line from the laser
to that height. They should then find the measure of the angle of the laser with a
protractor.
Ask the students: “What if we did not have a protractor? How could we find the angle?”
The equation for finding the angle, given the height and the distance from the wall, is:
15 cm
tan  
10 cm
We can solve for the angle by taking the inverse tangent of 1.5.
When students do this, they will have to use the “2nd”, “shift” or “inv” button on their
calculator, then press “tan”. 
Depending on the type of calculator, they may have to
divide 15 by 10 first, then find the arctan, or they can type 2nd, tan, 15 divided by 10
equals. Also be aware that some calculators, when doing this second kind of syntax,
immediately insert a parenthesis after the arctan, in which case typing arctan(15/10) will
provide the correct answer; other calculators do not, in which case students will need to
place the 15/10 in parentheses, or the order of operations will calculate (arctan15)/10.
Have students check their calculations against their measurements so that they can
connect what they are doing with their calculators to the activity.
16
Directions:
a) Measure a height on the wall
of 15 cm, then draw the line from
the laser to that height.
b) Find the measure of the angle
of the laser with a protractor.
The equation for finding the
angle, given the height and the
distance from the wall, is:
15 cm
tan  
10 cm
We can solve for the angle by
taking the inverse tangent of 15
divided by 10, or 1.5.

Draw the line for each of the
following heights, and use the
inverse tangent to calculate the
measure of the angle. Then
check your answers by
measuring the angle with a
protractor.
c)
The height is 1 cm.
wall
d)
The height equals the
distance to the wall.
e)
The height is twice the
distance to the wall.
f)
The height is half the
distance to the wall.
laser
10 cm
17
The equation for finding the angle, given the height and the distance from the wall, is:
15 cm
tan  
10 cm
We can solve for the angle by taking the inverse tangent of 1.5.
Answer drawings should look like this:

e) 20cm
tan =2
m=63.43°
a) 15 cm
tan =1.5
m=56.31°
wall
d) 10 cm
tan =1
m=45°
f) 5 cm
tan =0.5
m=26.57°
c) 1 cm
tan =0.1
m=5.71°
laser
10 cm
18
Part 4 – Graphing the tangent function
Worksheet 3
Use the data that you gathered in Part 2 to show graphically what is happening when
the angle of the laser increases.
Plot the angle (in degrees) on the x – axis, and the height of the wall on the y – axis.
When you have plotted all of your points, connect the dots to create a smooth curve.
18
16
14
12
Height
of
image
(cm)
10
8
6
4
2
0
0°
10°
20°
30°
40°
50°
Elevation of laser
60°
70°
80°
90°
At what point does the graph of tangent start? Why?
Why is the graph of the tangent not a straight line?
What is the behavior of the graph as it gets closer to 90 on the x – axis?
Will the graph ever cross the line x = 90? If so, where; if not, why not?
19
Worksheet 3 answers
18
16
14
12
Height
of
image
(cm)
10
8
6
4
2
0
0°
10°
20°
30°
40°
50°
Elevation of laser
60°
70°
80°
90°
Where does the graph of tangent start? Why?
The graph starts at (0, 0), because if the angle is 0, then the height is also 0, and so
tan  =(0/10).
Why is the graph of the tangent not a straight line?
As the angle increases, the height of the image increases at a faster rate.
Will the graph ever cross the line x = 90? If so, where; if not, why not?
No, the graph will never cross x = 90, because when the angle is 90, the line points
straight up and does not intersect the wall, so there is no ratio to be made.
20
Part 5 – Negative angles and tangents
Worksheet 4
Reflection: The graph we have just made starts at (0, 0 cm); what possible meanings
could there be in our laser – scenario for an angle of -10?
At the South Rim, the width of the Grand Canyon is approximately 16 kilometers. What
would the depth of our laser image be, if the laser were rotated down by -5? Try
calculating the length of the image using the negative angle in the tangent function, and
give a plausible interpretation of your answer.
laser
16 km
-5°
image
(not to scale)
Grand Canyon
wall
21
Go back to your data table in part 2 and calculate the heights of the negatives of the
angles; we can then plot them on a coordinate plane as before, getting a more complete
idea of the tangent function. What kind of symmetry does the graph show? What
happens to the graph when the angle measure gets closer to -90?
18
16
14
12
10
8
6
4
2
-90° -80° -70° -60° -50° -40° -30° -20° -10°
0°
10° 20°
30° 40° 50°
60° 70° 80° 90°
-2
-4
-6
-8
-10
-12
-14
-16
-18
22
Worksheet 4 answers
Reflection: The graph we have just made starts at 0; what possible meanings could
there be in our laser-scenario for an angle of -10?
The negative angle means we are rotating the laser down from the line at 0 .
At the South Rim, the width of the Grand Canyon is approximately 16 kilometers. What
would the depth of our laser image be, if the laser were rotated down by -5? Try
calculating the length of the image using the negative angle in the tangent function, and
give a plausible interpretation of your answer.
laser
16 km
-5°
image
Grand Canyon
wall
(not to scale)

Answer: tan -5 = -0.09, solving for the image depth gives us
16km (tan  5)  1.40 km, or 1.40 km lower than the laser.
The graph should look like this:
18
The graph of the tangent function
16
shown has rotational symmetry of
14
180 about the origin, meaning that
12
if you replace x with –x, the
10
corresponding y – value will be –y;
8
or tan (–x) = – tan x.
6
As x approaches –90 from the right,
y decreases asymptotically.
4
2
-90° -80° -70° -60° -50° -40° -30° -20° -10°
0°
10° 20°
30° 40° 50°
60° 70° 80° 90°
-2
-4
-6
-8
-10
-12
-14
-16
-18
23
IV.
APPLICATION:
Part 6 – More complicated tangent problems
Teacher notes: This should be a teacher - led discussion on how to break figures down
into sub-problems, and redraw parts of figures to provide clarity.
Example 1—finding a missing side
Instead of having a laser image covering the wall all the way from the base of the wall,
we might want an image that only covers part of the wall (from point A to point B below).
A
image
(h)
B
wall
14°
laser
38°
32 m
P
C
Here, we know the distance of the laser to the wall (32 m), and we know the angle that
the top and bottom of the image make with the ground (38 and 52); we want to
determine the length of the line segment AB.
A good way to approach the problem is to separate the two triangles we have, and
solve for their heights separately:
A
B
x
y
laser
laser
52°
P
Solve for x:
32 m
C
38°
P
32 m
C
Solve for y:
x
y
tan 38 
32 m
32 m
x = 1.280(32 m)
y = 0.7813(32 m)
x = 40.93 m
y = 25.00 m
Then the height h of the image is x – y, or 40.93 m – 25.00 m, or 15.93 m.

tan 52 

24
Example 2 – finding a missing angle
Now we want to know the measure of the angle between the top and bottom lasers, or
in the figure below, the measure of APB.
A
image
10 m
B
120 m
wall
laser
P
100 m
C
In this figure, the height of the image is 10 m, while the image itself starts 120 m above
the floor.
Steps: Calculate the measures of BPC and APC, then subtract in order to figure out
the measure of APB.
120m 
120m
Answers: tanBPC 
, so BPC = tan1
 50.19;
100m
100m 
130m 
130m
tanAPC 
, so APC tan1
 = 52.43;
100m
100m 

APB
= 52.43 – 50.19 = 2.24.



25
Example 3—finding a missing adjacent side length
Teacher’s note: This example is very complicated in terms of the algebraic solution.
The values for tan 50 and tan 55 have been left in that form until the end of the
problem, but they could very easily be calculated at the beginning, especially for
students who are not used to working abstractly.
A
10m
B
h
5°
laser
50°
P
d
C
If we know the two angles of elevation and the height of the image, we can determine
both the height of the image above the floor and the distance between the wall and the
laser:
h + 10
h
We know tan 50  , and tan 55 
. We can solve both of these for d:
d
d
h
h + 10
d=
and d =
.
tan 50
tan 55
h
h  10

 that means

So
.
tan 50 tan 55
Wecan cross-multiply:
h tan 55  (h 10) tan 50

Distribute
htan 55  htan 50 10tan 50
Subtract
h tan 55  h tan 50  10 tan 50

Factor
h (tan 55 tan 50)  10 tan 50

10 tan50
h 
 50.41m
Divide
(
tan
55

tan50)

50.41m
d
 42.30m
Then 
tan 50


26
Part 7 – Problem solving and drawing auxiliary lines
Teacher notes: Again, this should be teacher – led.
Example 1: In many cases, we will need to draw extra lines and right triangles in order
to use the tangent angle.
108°
8m
Find the area of the regular pentagon, given that the area of a triangle is
1
bh , if b is the
2
base of the triangle and h is its height.
First, divide the pentagon into 5 triangles by connecting segments from the center to
each vertex (below left). Notice that the measure of the angle between a side and the

line to the center is 54. (We can derive the angle 108 by finding
one central angle;
360/5 = 72; then the two base angles of the isosceles triangle add to 108.)
54°
54°
h
8m
54°
4m
In order to find the area of one of the triangles, we need the height, so draw a
perpendicular from the center to the midpoint of one side (above right).
h
Then, to find the height, we use tan 54 =
, so h = (4 m)(1.38) = 5.51 m.
4m
1
Then the area of one triangle is A  ( 8m)( 5.51m) = 22.02 m2, and the area of the
2
2) = 110.11m2.
entire pentagon is 5(22.02
m


27
Example 2: Jane is trying to measure the height of a waterfall. Her friends Nicole and
Evan have helped her measure the angles from the top and bottom of the fall, and she
knows the distance to the waterfall at the base of the cliff. How can she calculate the
height?
In order to use the tangent ratio
Nicole
here, we need to have a right
triangle. Draw a perpendicular line
62°
from Jane’s vertex to the waterfall,
as shown below:
Jane
waterfall
To use these right triangles, we next
need to calculate the unknown
angles in the figure, using the fact
that the angles have to sum to 180.
cliff
41°
196 m
Evan
Nicole
62°
Jane
waterfall
cliff
41°
Evan
196 m
28
Nicole
62°
a
28°
196 m
waterfall
Jane
41°
b
cliff
49°
41°
Evan
196 m
Now we can use the tangent ratio to solve for the lengths a and b in the picture above:
a
, so a  196 m tan 28 = 104.22 m, and
196m
b
tan 41 
, so b  196 mtan 41 = 170.38 m.
196m
 of the waterfall is a + b = 274.60 m.
Then the height
tan 28 



29
V.
ASSESSMENT:
Post-assessment
1) Solve for the measure of B:
C
8.5
16.2
A
B
a) 27.69 (correct answer)
b) 31.65 (used inverse sine of (8.5/16.2))
c) 58.35 (used inverse cosine of (8.5/16.2))
d) 62.31 (used inverse cotangent of (8.5/16.2))
(CRS target: 5.4; recognize and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems.)
2) A ray starts at the origin and passes through the point (7, 24).
y
(7, 24)
x
Calculate the angle between the x – axis and the ray.
a) 16.26 (used inverse cotangent)
b) 16.96 (used inverse sine of 7/24)
c) 73.04 (used inverse cosine of 7/24)
d) 73.74 (correct answer)
(CRS target: 5.4; recognize and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems.)
30
3) The figure below shows a lighthouse, with fog 8 kilometers away. The beam of light
is currently at the point A, and the light spins clockwise at a rate of 1 per second.
fog
F
10 km
A
B
8 km
G
lighthouse
How many seconds did it take for the beam to get from point F to point A?
a) 1.25 seconds (did not take inverse tangent of 10/8)
b) 38.66 seconds (found inverse tangent of 8/10 instead of 10/8)
c) 51.34 seconds (correct answer)
d) 53.13 seconds (found the inverse sine of 8/10)
(CRS target: 5.4; recognize and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems.)
31
4) The figure below shows a lighthouse, with fog 8 kilometers away. The beam of light
is currently at the point A, and the light spins clockwise at a rate of 1 per second.
fog
F
8 km
A
B
48°
lighthouse
If the point B is where the beam of light is five seconds later, calculate the distance from
A to B.
a) 0.70 km (student found 8*tan 5))
b) 1.73 km (correct answer)
c) 9.58 km (student found 8*tan 48 + 8*tan 5)
d) 10.62 km (did not subtract length FA from length FB)
(CRS target: 5.4; recognize and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems.)
32
5) The figure below shows a lighthouse, with fog 8 kilometers away. The beam of light
is currently at the point A, and the light spins clockwise at a rate of 1 per second.
fog
F
10 km
A
B
8 km
G
lighthouse
If the point B is 5 kilometers away from A, how many seconds does it take for the light
beam to move from A to B?
a) 10.59 seconds (correct answer, tan-1(15 / 8)  tan-1(10 / 8) )
b) 25.67 seconds (found AGF, then multiplied answer by 0.5)
c) 61.93 seconds (found BGF)
d) 77.01 seconds (found AGF, then multiplied answer by 1.5)
(CRS target: 5.4; recognize 
and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems.)
33
6) In the figure below, the lengths of QR and RT are both 10, the length of PT is 25.
Q
10
R
x
10
P
25
T
Calculate x, the measure of QPR.
a) 38.66 (student found inverse tangent of 20/25)
b) 21.80 (student found inverse tangent of 10/25)
c) 19.33 (student found half of inverse tangent of 20/25)
d) 16.86 (correct answer; student found inverse tangent of 20/25 and inverse
tangent of 10/25, and subtracted.)
(CRS target: 5.4; recognize and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems.)
34
7) The figure below shows the graph of the line y = x. The slope of this line is 1, and
the angle that line makes with the x – axis is 45.
8
y=x
6
4
2
45.0°
-8
-6
-4
-2
2
4
6
8
-2
-4
-6
-8
Calculate the slope of a line that makes an angle of 15 with the x – axis.
a) 0.27 (correct answer)
b) 0.33 (divided the slope of the line by 3)
c) 3.00 (multiplied the slope of the line by 3)
d) 3.73 (found the slope of a line making a 15 angle with the y – axis)
(CRS target: 5.4; recognize and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems.)
35
8) In the figure below, calculate the length AB, the height of the image projected onto
the wall from the laser.
A
image
(h)
B
wall
9°
laser
33°
P
20 m
C
a) 3.17 m (found 20*tan 9)
b) 5.02 m (correct answer)
c) 14.84 m (found 20*tan 42 - 20*tan9)
d) 18.01 m (found AC)
(CRS target: 5.4; recognize and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems.)
36
9) In the figure below, a telephone pole is 25 meters from a house. The top of the pole
is 13 meters high, and there is a wire that goes from the top of the pole to the house,
where it connects 9 meters off the ground.
wire
pole

height
13 m
house
height
9m
25 m
What is the measure of the angle  in the figure?
a) 7.68 (student found the inverse tangents of 13/25 and 9/25 and subtracted)
b) 9.09 (correct answer)
c) 27.47 (student found the inverse tangent of 25/13)
d) 80.91 (student found the inverse tangent of 25/4)
(CRS target: 5.4; recognize and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems.)
37
Short answer (2 points)
10) In the figure below, the length of BC is 12 units and the length of DC is 9 units. The
measure of BAC is two – thirds the measure of BDC. Right angles are indicated.
B
12
2
3
x
A
x
D
9
C
Calculate the value of x and the length AD. Be sure to:
 Apply the tangent ratio in your solution
 Support both answers with the steps showing your calculations
Measure of x: _______
Length of AD: _______
38

2-point response: The student shows understanding of recognizing and applying the
tangent right triangle relationship to solve problems by doing the following:
 Uses arc tan to find the angle x = 53.13
 Uses tan to find the length AD = 7.87 units.
1-point response: The student shows some understanding of recognizing and applying
the tangent right triangle relationship to solve problems by doing one of the following:
 Uses arc tan to find the angle x = 53.13
 Uses arc tan to find an angle x and uses that value in the solution for length AD
0-point response show little or no understanding of recognizing and applying the
tangent right triangle relationship to solve problems.
Example solution:
12 
12
2
tan x =
, so x = tan -1   53.13 and x  35.42.
9
3
 9 
12 


12
Then tan 35.42    , so AC  
 = 16.87, and AD = 16.87 – 9 = 7.87
AC 
tan 35.42  


(CRS target: 5.4; recognize and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems.)


39
11) In the figure below, Alice is trying to figure out the height of the cliff across the
canyon from her. She knows that the canyon itself is 256 meters across, and Bob and
William have measured angles for her as indicated.
W
William
55°
Alice
A
cliff
256m
40°
Bob
B
Calculate the length of WB. Be sure to:



Draw any auxiliary lines and mark the measurements of angles you make
Apply the tangent ratio in your solution
Support your answers with the steps showing your calculations
Length of WB: _______
40
2-point response: The student shows understanding of analyzing a situation and
describe the problems to be solved, and recognizing and applying the tangent right
triangle relationship to solve problems by doing the following:
 Draws a perpendicular line from point A to side WB, making 2 right triangles
 Determines the missing angles from point A in the two triangles (35 and 40)
 Uses tan to find the lengths of the opposite sides of the right triangles (179.25 m
and 214.81 m)
 Determines the total length of WB = 394.06
1-point response: The student shows some understanding of analyzing a situation and
describe the problems to be solved, and recognizing and applying the tangent right
triangle relationship to solve problems by doing two of the following:
 Draws a perpendicular line from point A to side WB, making 2 right triangles
 Determines the missing angles from point A in the two triangles (35 and 40)
 Uses tan to find the lengths of the opposite sides of the right triangles (179.25 m
and 214.81 m)
 Determines the total length of WB = 394.06
0-point response show little or no understanding of analyzing a situation and
describing the problems to be solved, or recognizing and applying the tangent right
triangle relationship to solve problems.
Example solution:
Draw the perpendicular from A to WB,
W William
and find the measures of the missing
a
55°
angles. Then tan 35 =
,
256 m
a
so a = 256 tan 35 = 179.25 m.
b
35°
Also tan 40 =
,
Alice A
256
m

40°
cliff
So b = 256 tan 40 = 214.81m.

Then the height of the cliff is
a + b = 394.06 m.

b

50°
256m
40°
Bob
B
(CRS targets: 5.4: Recognize and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems); 1.1 : Analyze a situation and describe the
problems to be solved.)
41
12) In the figure below, the length of QS is 1 unit and the length of PS is 2 units. Right
angles are indicated.
Q
1

P
2
R
S
x
Calculate the value of  and the length of SR. Be sure to:


Apply the tangent ratio in your solution
Support both answers with the steps showing your calculations
42




Measure of : _______ Length of SR: _______
2-point response: The student shows understanding of recognizing and applying the
tangent right triangle relationship to solve problems by doing the following:
 Uses arc tan to find the angle theta = 26.57
 Uses tan to find the length SR = 1/2 unit.
1-point response: The student shows some understanding of recognizing and applying
the tangent right triangle relationship to solve problems by doing one of the following:
 Uses arc tan to find the angle theta = 26.57
 Uses arc tan to find an angle theta and uses that value in the solution for length
SR
0-point response show little or no understanding of recognizing and applying the
tangent right triangle relationship to solve problems.
Example solution 1:
1
1
tan   , so  = tan1( ) , and  = 26.57
2
2
1
QRP = 90 – 26.57 = 63.43, so tan (64.43 ) 
x
1
1
x = 

tan (63.43) 2

Example solution 2:
1
1
tan   , so  = tan1( ) , and  = 26.57
2
2
PQS = 90 – 26.57 = 63.43, and RQS =90 – 63.43  = 26.57,
x
1
tan (26.57 )  , so x =
.
1
2

(CRS target: 5.4; recognize and apply the basic right-angle triangle trigonometric
relationship (tangent) to solve problems.)

43
13) In the figure below, ray PT is tangent to the circle C at the point T. Right angles are
indicated. The area of a circle is A    r 2 .

T
r
8m

P
C
a) If the angle  = 22, find the radius of the circle, and the area.
b) Find the measure of the angle  that would double the area of the circle.
Be sure to:


Apply the tangent ratio in your solution
Support both answers with the steps showing your calculations
Length of radius:___________ Area of circle:___________ Measure of : _______
44
2-point response: The student shows understanding of recognizing and applying the
tangent right triangle relationship to solve problems by doing the following:
 Uses tan to find the radius r = 3.23 m, and the area 32.82 m2
 Uses arc tan to find the angle  = 29.74.
1-point response: The student shows some understanding of recognizing and applying
the tangent right triangle relationship to solve problems by doing one of the following:
 Uses tan to find the radius r = 3.23 m, and the area 32.82 m2
 Uses tan to find a radius and the associated area, and uses that value in the
solution for angle .
0-point response show little or no understanding of recognizing and applying the
tangent right triangle relationship to solve problems.
Example solution:
r
, so r  8 m tan 22 = 3.23m.
tan 22 
8m
Then the area of the circle is   (3.23 m)2 = 32.82 m2.
The circlewith double the area has an area of 65.64 m2. To find the radius, we have
A    R2 , or 65.64 m2    R2 , so R2 = 20.89 m2 and R = 4.57 m.

4.57 m
Then the tangent of the angle we are looking for is tan  
,
8m
4.57 m 
so  tan1
 =29.74.
 8 m 




45
Extended response (4 points)
14) A laser is aimed at a wall 24 meters away.
a) Find the angle that the laser makes with the ground if the height of the laser on the
wall is 5 m above the ground. Then complete the table for the missing values of the
angles.
b) At what angle would the laser have to be to get a height of 24 meters on the wall?
c) Graph the results that you have calculated, and describe the behavior of the graph
as the angle gets closer to 90.
In writing your answer, be sure to:
 Apply the tangent ratio in your solution.
 Support all three answers with the steps showing your calculations and a
diagram of the situation.
 For part c, be sure to show the appropriate scale on the graph and use your
graph to support your answer.
height
of wall
angle of
laser
Diagram
5m
10 m
15 m
20 m
46
(continue your answer on the next page…)
90°
80°
70°
60°
Elevation
of laser
50°
40°
30°
20°
10°
0
Height of wall (m)
47
answers: part a)
height
angle of
of wall
laser
5m
11.77
10 m
22.62
15 m
32.01
diagram
5m
5m
wall
5m
laser
5m
20 m
39.81
24 m
b) At what angle would the laser have to be to get a height of 24 meters on the wall?
45
c)
90°
80°
70°
60°
Elevation
of laser
50°
40°
30°
20°
10°
0
24m
Height of wall (m)
As the height of the wall increases, the angle of the laser approaches 90. The graph
should start at (0, 0) and be increasing and asymptotic as it approaches 90.
(CRS target 8.6; recognize, analyze, interpret, and model with trigonometric functions.)
48
A 4 – point response: The student shows understanding of recognizing, analyzing,
interpreting, and modeling with trigonometric functions by earning 7 or 8 of the following
value points:
 Calculates the angles of the laser for each height in the table (2 points)
 Draws a correct diagram for at least one of the heights (1 point)
 Writes the angle of the laser in part b as 45 (1 point)
 Creates an appropriately labeled graph with at least 5 correct points that shows
asymptotic behavior as the graph approaches 90 (2 points)
 Describes the behavior of the graph as it approaches 90 (2 points)
Note: Allow for one calculation error and any answers that follow from that error.
A 3 – point answer: The student earns 5 or 6 value points.
A 2 – point answer: The student earns 3 or 4 value points.
A 1 – point answer: The student earns 1 or 2 value points.
A 0 – point answer shows little or no understanding of recognizing, analyzing,
interpreting, and modeling with trigonometric functions by earning 0 or 1 value points.
49
15) The regular octagon below has side lengths of 10 m each, and an interior angle of
135 at each vertex.
135°
10 m
a) Divide the octagon into 8 congruent triangles and use the tangent ratio to calculate
the area of the octagon.
b) Use the method that you have just completed in part a) to calculate the area of a
regular polygon with 100 sides that has side lengths of 10 m each, and an interior angle
of 176.4.
In writing your answer, be sure to:
 Apply the tangent ratio in your solution.
 Support all answers with the steps showing your calculations and diagrams of the
situation.
 Label all diagrams appropriately.
(continue your answer on the next page…)
50
answers: part a)
First we divide the octagon into 8
congruent isosceles triangles, with base
angles of 67.5, a base of 10 m, and a
height h. We can then find the height in
one of two ways:
h
tan 67.5  = ,
5
so h = 5 tan 67.5 = 12.07m; or

22.5°
h

67.5°
5m

tan 22.5  =
h=
5
, so h tan 22.5 = 5 and
h
5
  = 12.07m.
tan 22.5

1
Then the area of one of the isosceles triangles is  bh , which is
2
1
(10m)(12.07m) = 60.36 m2 .
2
Thus, the area of the whole octagon is 860.36 m2  484.84 m2



51
Part b)
1.8°
h

88.2°
5m


It will not be possible to draw an accurate 100-gon, but the
triangle above shows one of the isosceles triangles that
would be made (not to scale). The base angle of the triangle
is 88.2, with a base of 10 m and a height of h. Then we
calculate the area of that triangle in the same way as before:
h
tan 88.2  = , so h = 5 tan 88.2 = 159.10 m, or
5
5
5
tan 1.8  = , so h tan 1.8 = 5 , and h =
= 159.10 m.
h
tan 1.8 

Then the area of one of these isosceles triangles is


1
(10m)(159.10m) = 795.51 m2
2
and the total area of the 100-gon is

100 795.51 m2  79951.29 m2.
A 4 – point response: The student shows understanding of abstracting mathematical
models from word 
problems, geometric problems, and applications by earning 7 or 8 of
the following value points:
 Draws a correct diagram for the triangles in the octagon (1 point)
 Uses tangent to calculate the height of the isosceles triangle (1 point)
 Calculates the area of the isosceles triangle of the octagon (1 point)
 Calculates the area of the octagon (1 point)
 Draws a correct diagram for the triangles in the 100-gon (1 point)
 Uses tangent to calculate the height of the isosceles triangle (1 point)
 Calculates the area of the isosceles triangle of the 100-gon (1 point)
 Calculates the area of the 100-gon (1 point)
Note: Allow for one calculation error and any answers that follow from that error.
A 3 – point answer: The student earns 5 or 6 value points.
A 2 – point answer: The student earns 3 or 4 value points.
A 1 – point answer: The student earns 1 or 2 value points.
A 0 – point answer shows little or no understanding of recognizing, analyzing,
interpreting, and modeling with trigonometric functions by earning 0 or 1 value points.
52
CRS targets 3.4 – Abstract mathematical models from word problems, geometric
problems, and applications.
53
VI.
EXTENSIONS:
One possible avenue to pursue after this lesson is using the unit circle to generate the
tangent line.
On unit circle, the tangent ratio is shown by the length of the tangent to the circle at the
point (1,0).
We can think of tangent not just as a ratio in a triangle but as a height (or depth) on the
line x = 1 (or -1).
P
tan 
1
sin 

cos 
Q
Develop symmetric properties of tangent, as pointing up to Quadrant I and down into
Quadrant III, and symmetric about origin.
Connect to negative angle measure in Quadrant IV…
For figuring out angles in Quadrant II and Quadrant III, we need to think of theta as the
reference angle… so that for Quadrant II, we take 180 – , while for Quadrant III we
take  – 180
54
1.2
1
0.8
0.6
0.4
tan 
0.2

-2
-1.5
-1
-0.5
0.5
1
1.5
2
-0.2
tan 
-0.4
-0.6
-0.8
-1
-1.2
1.4
1.2
1
0.8
for negative values of
 (between 0° and
90°),
tan  is negative
0.6
tan 
0.4
0.2
-2
-1.5
-1
-0.5

0.5
1
1.5
2
-0.2
-0.4
tan 
-0.6
-0.8
-1
55
VII. STUDENT WORKSHEETS AND ASSESSMENTS (BLANK)
Pre-assessment
1) Solve for x:
12
33°
x
a) 7.79
b) 14.31
c) 18.48
d) 22.03
2) Solve for the measure of A:
B
18
A
14
C
a) 1.3
b) 22.8
c) 37.9
d) 52.1
56
3) Solve for the length of QR:
Q
65°
R
P
10
a) 5.73
b) 7.00
c) 9.06
d) 21.44
4) Calculate the tangent of B.
A
10
6
C
a)
b)
c)
d)
8
B
0.60
0.75
0.80
1.33
57
5) A ray starts at the origin and passes through the point (12, 5). Solve for the angle
between the x – axis and the ray.
(12, 5)
(0, 0)
a)
b)
c)
d)
67.38
65.38
24.62
22.62
58
Worksheets
Worksheet 1
Angle
Height of image
(in cm)
0
0
Difference in height
(subtract previous
height)
------
10
20
30
40
50
60
70
80
85
89
When you have finished your table:
1) Describe the behavior of the height as the angle approaches 90.
2) Does the height increase at a steady rate?
3) What happens to the intersection of the laser and the wall when the angle of the
laser is 90?
4) What happens to the differences in height as the angle of the laser gets larger?
59
Directions:
a) Measure a height on the wall
of 15 cm, then draw the line from
the laser to that height.
b) Find the measure of the angle
of the laser with a protractor.
The equation for finding the
angle, given the height and the
distance from the wall, is:
15 cm
tan  
10 cm
We can solve for the angle by
taking the inverse tangent of 15
divided by 10, or 1.5.

Draw the line for each of the
following heights, and use the
inverse tangent to calculate the
measure of the angle. Then
check your answers by
measuring the angle with a
protractor.
c)
The height is 1 cm.
wall
d)
The height equals the
distance to the wall.
e)
The height is twice the
distance to the wall.
f)
The height is half the
distance to the wall.
laser
10 cm
60
Worksheet 3
Use the data that you gathered in Part 2 to show graphically what is happening when
the angle of the laser increases.
Plot the angle (in degrees) on the x – axis, and the height of the wall on the y – axis.
When you have plotted all of your points, connect the dots to create a smooth curve.
18
16
14
12
Height
of
image
(cm)
10
8
6
4
2
0
0°
10°
20°
30°
40°
50°
Elevation of laser
60°
70°
80°
90°
At what point does the graph of tangent start? Why?
Why is the graph of the tangent not a straight line?
What is the behavior of the graph as it gets closer to 90 on the x – axis?
Will the graph ever cross the line x = 90? If so, where; if not, why not?
61
Worksheet 4
Reflection: The graph we have just made starts at (0, 0 cm); what possible meanings
could there be in our laser – scenario for an angle of -10?
At the South Rim, the width of the Grand Canyon is approximately 16 kilometers. What
would the depth of our laser image be, if the laser were rotated down by -5? Try
calculating the length of the image using the negative angle in the tangent function, and
give a plausible interpretation of your answer.
laser
16 km
-5°
image
(not to scale)
Grand Canyon
wall
62
Go back to your data table in part 2 and calculate the heights of the negatives of the
angles; we can then plot them on a coordinate plane as before, getting a more complete
idea of the tangent function. What kind of symmetry does the graph show? What
happens to the graph when the angle measure gets closer to -90?
18
16
14
12
10
8
6
4
2
-90° -80° -70° -60° -50° -40° -30° -20° -10°
0°
10° 20°
30° 40° 50°
60° 70° 80° 90°
-2
-4
-6
-8
-10
-12
-14
-16
-18
63
Post-assessment
1) Solve for the measure of B:
C
8.5
16.2
A
B
a)
b)
c)
d)
27.69
31.65
58.35
62.31
2) A ray starts at the origin and passes through the point (7, 24).
y
(7, 24)
x
Calculate the angle between the y – axis and the ray.
a)
b)
c)
d)
16.26
16.96
73.04
73.74
64
3) The figure below shows a lighthouse, with fog 8 kilometers away. The beam of light
is currently at the point A, and the light spins clockwise at a rate of 1 per second.
fog
F
10 km
A
B
8 km
G
lighthouse
How many seconds did it take for the beam to get from point F to point A?
a) 1.25 seconds
b) 38.66 seconds
c) 51.34 seconds
d) 53.13 seconds
65
4) The figure below shows a lighthouse, with fog 8 kilometers away. The beam of light
is currently at the point A, and the light spins clockwise at a rate of 1 per second.
fog
F
8 km
A
B
48°
lighthouse
If the point B is where the beam of light is five seconds later, calculate the distance from
A to B.
a) 0.70 km
b) 1.73 km
c) 9.58 km
d) 10.62 km
66
5) The figure below shows a lighthouse, with fog 8 kilometers away. The beam of light
is currently at the point A, and the light spins clockwise at a rate of 1 per second.
fog
F
10 km
A
B
8 km
G
lighthouse
If the point B is 5 kilometers away from A, how many seconds does it take for the light
beam to move from A to B?
a)
b)
c)
d)
10.59 seconds
25.67 seconds
61.93 seconds
77.01 seconds
67
6) In the figure below, the lengths of QR and RT are both 10, the length of PT is 25.
Q
10
R
x
10
P
25
T
Calculate x, the measure of QPR.
a)
b)
c)
d)
38.66
21.80
19.33
16.86
68
7) The figure below shows the graph of the line y = x. The slope of this line is 1, and
the angle that line makes with the x – axis is 45.
8
y=x
6
4
2
45.0°
-8
-6
-4
-2
2
4
6
8
-2
-4
-6
-8
Calculate the slope of a line that makes an angle of 15 with the x – axis.
a)
b)
c)
d)
0.28
0.33
3.00
11.43
69
8) In the figure below, find the length AB, the height of the image projected onto the
wall from the laser.
A
image
(h)
B
wall
9°
laser
33°
P
a)
b)
c)
d)
20 m
C
3.17 m
5.02 m
14.84 m
18.01 m
70
9) In the figure below, a telephone pole is 25 meters from a house. The top of the pole
is 13 meters high, and there is a wire that goes from the top of the pole to the house,
where it connects 9 meters off the ground.
wire
pole

height
13 m
house
height
9m
25 m
What is the measure of the angle  in the figure?
a) 7.68
b) 9.09
c) 27.47
d) 80.91
71
Short answer (2 points)
10) In the figure below, the length of BC is 12 units and the length of DC is 9 units. The
measure of BAC is two – thirds the measure of BDC. Right angles are indicated.
B
12
2
3
x
A
x
D
9
C
Calculate the value of x and the length AD. Be sure to:
 Apply the tangent ratio in your solution
 Support both answers with the steps showing your calculations
Measure of x: _______
Length of AD: _______
72
11) In the figure below, Alice is trying to figure out the height of the cliff across the
canyon from her. She knows that the canyon itself is 256 meters across, and Bob and
William have measured angles for her as indicated.
W
William
55°
Alice
A
cliff
256m
40°
Bob
B
Calculate the length of WB. Be sure to:



Draw any auxiliary lines and mark the measurements of angles you make
Apply the tangent ratio in your solution
Support your answers with the steps showing your calculations
Length of WB: _______
73
12) In the figure below, the length of QS is 1 unit and the length of PS is 2 units. Right
angles are indicated.
Q
1

P
2
R
S
x
Calculate the value of  and the length of SR. Be sure to:


Apply the tangent ratio in your solution
Support both answers with the steps showing your calculations
Measure of : _______
Length of SR: _______
74
13) In the figure below, ray PT is tangent to the circle C at the point T. Right angles are
indicated. The area of a circle is A    r 2 .

T
r
8m

P
C
a) If the angle  = 22, find the radius of the circle, and the area.
b) Find the measure of the angle  that would double the area of the circle.
Be sure to:


Apply the tangent ratio in your solution
Support both answers with the steps showing your calculations
Length of radius:___________ Area of circle:___________ Measure of : _______
75
Extended response (4 points)
14) A laser is aimed at a wall 24 meters away.
a) Find the angle that the laser makes with the ground if the height of the laser on the
wall is 5 m above the ground. Then complete the table for the missing values of the
angles.
b) At what angle would the laser have to be to get a height of 24 meters on the wall?
c) Graph the results that you have calculated, and describe the behavior of the graph
as the angle gets closer to 90.
In writing your answer, be sure to:
 Apply the tangent ratio in your solution.
 Support all three answers with the steps showing your calculations and a
diagram of the situation.
 For part c, be sure to show the appropriate scale on the graph and use your
graph to support your answer.
height
of wall
angle of
laser
Diagram
5m
10 m
15 m
20 m
76
(continue your answer on the next page…)
90°
80°
70°
60°
Elevation
of laser
50°
40°
30°
20°
10°
0
Height of wall (m)
77
15) The regular octagon below has side lengths of 10 m each, and an interior angle of
135 at each vertex.
135°
10 m
a) Divide the octagon into 8 congruent triangles and use the tangent ratio to calculate
the area of the octagon.
b) Use the method that you have just completed in part a) to calculate the area of a
regular polygon with 100 sides that has side lengths of 10 m each, and an interior angle
of 176.4.
In writing your answer, be sure to:
 Apply the tangent ratio in your solution.
 Support all answers with the steps showing your calculations and diagrams of the
situation.
 Label all diagrams appropriately.
(continue your answer on the next page…)
78
79