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Unit 3 – Right Triangle Trigonometry - Classwork
We have spent time learning the definitions of trig functions and finding the trig functions of both quadrant and
special angles. But what about other angles? To understand how to do this, and more importantly, why we do
it, we introduce a concept called the unit circle. A unit circle is a circle whose radius is one.
To the left is a unit circle. The angle
" is drawn in the first quadrant but could be
drawn anywhere. Suppose
. If we were to find
sin 40 ° , we know that it
would be defined as
y
= y. So when we take
sin 40 °, we are finding the height
of the triangle in a unit circle. The same argument holds when we take
cos 40 ° …
we are actually finding the x variable in a unit circle. When we take
tan 40 °, we
are finding the ratio of y to x in a unit circle.
On your calculator, be sure you are in Degree Mode and set your decimal accuracy
to FLOAT. Use your calculator to find
sin 40 ° and
cos 40 °. Remember what it is
you are finding: the y and x variables in the triangle above. And since
x = cos 40 ° and y = sin 40 °
, let us show that the Pythagorean theorem holds in this
triangle based on the unit circle.
Taking trig functions on the calculator is straightforward: type in the trig function (you will get a left
parentheses) and the angle. You do not need to complete the parentheses. Press ENTER and out it comes.
Although we can get extreme accuracy, we will find that four decimal places is usually enough. So set your
calculator to 4 decimal places. Remember that angles are assumed to be in radians unless in degree format.
Example 1) Find the following:
a)
sin 29 ° =. b)
cos 131 ° = ". c)
tan
If angles are input with more accuracy, it is assumed that they are in decimal degrees. Note that parentheses can
be used to make the problems clearer in intent.
Example 2) Find the following:
a)
tan12.8° =. b)
sin "32.35°
c)
cos 0.724°
If trig functions of angles that are in degree-minute-second form, use the Angle menu to
input them. Remember that seconds are input with ALPHA +.
cos 38 ° 4 0 " 2 9 " "would be
input to the calculator thusly:
Example 3) Find the following:
a)
sin 82 ° 1
2 = .9907 b)
cos 126 ° 4
3 = #.5978 c)
tan " 8 ° 5
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Note that there are no keys for the csc, sec, or cot functions on your calculator. To
find them we have to use the fact that sin and csc functions are reciprocals of each
other, as are the cos and sec functions, and the tan and cot functions. There are three
ways to find, for example
csc 37 °. Take
sin 37 ° and then take its reciprocal or simply
finding 1/
sin 37 °. The screen on the right shows these two methods. You could also
find
sin 37 ° and then press the reciprocal key
x
" 1
Example 4) Find the following:
a)
csc 81 ° = 1.0125 b)
sec122° = "1.8871 c)
cot 34.2° = 1.
d)
sec 338.292° = 1. e)
cot 14 ° 2 9 " 3 6 " "= 3. f)
csc149° 5 0 " "= 1.
Many times, we want to reverse the process. We know the sine of an angle and we wish to find the angle itself.
To accomplish this, we use inverse trig functions or arc trig functions. These are found on your calculator
above the sin, cos, and tangent keys. We use the blue (
nd
) key to input them.
For instance, let us find the first quadrant angle whose sine is .7523. Note the screen on
the right. Our answer would be
48.79° (expressed in decimal degrees). If we wanted
our answer in degree – minute – second format, note how we would accomplish that by
using the Angle menu.
Example 5) Find the following (decimal degrees):
a)
sin
" 1
b)
arccos0.4231 = 64.9695° c)
tan
" 1
Example 6) Find the following (Degrees – minutes – seconds)
a)
arctan 4.002 = 75 ° 5 8 " 1 4 "" b)
sin
" 1
.0809 = 4 ° 3 8 # 2 5 # # c)
cos
" 1
Finally, if we wish to find an arccsc, arcsec, or arctan function, again, there is no one
keystroke that will give it to you. To find
csc
" 1
2.3552 , for instance, we must first take the
reciprocal of 2.3552, and then take the arcsin of that. On the right is the way you would
accomplish this (with the optional changing into degrees-minutes-seconds):
Example 7) Find the following (decimal degrees):
a)
sec
" 1
1.76 = 55.3765° b)
arccot 3.4221 = 16.2893° c)
csc
" 1
Example 8) Find the following (Degrees – minutes – seconds)
a)
arccsc 3.8621 = 15 °
3 b)
cot
" 1
5 c)
arcsec5.8621 = 80 ° 1
About errors:
Your calculator can take trig functions and arc trig functions to extreme accuracy. However, if you input the
problem into the calculator incorrectly, one of two things will happen. One – an error. Assuming you typed it
in the correct syntax, the calculator is telling you that it cannot perform the operation. This is actually good for
you. For instance, if you take
cos
" 1
1.4231 the calculator gives you a domain error because we know that the
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Example 9) Angle and hypotenuse
A = 21 ° a = 5.
B = 69 ° b = 13.
C = 90 ° c = 14
a = 14 sin 21 ° b = 14 cos 21 °
_______________________________________________________________________________________
Example 10) Angle and leg
A = 77.2° a = 29.
B = 12.8° b = 6.
C = 90 ° c = 30.
b = 29.5tan12.8° c =
sin 77.2°
______________________________________________________________________________________
Example 11) Angle and leg
A = 38 ° 1 2 " 4 4 " " a = 102.
B = 51 ° 4 7 " 1 6 " " b = 130.
C = 90 ° c = 165.
Note that we can find B:
b = 102.35tan 51 ° 4 7 " 1 6 " " c =
sin 38 ° 1 2 " 4 4 ""
______________________________________________________________________________________
Example 12) Leg and hypotenuse
A = 37.98° a = 8
B = 52.02° b = 10.
C = 90 ° c = 13
b = 169 " 64 A = sin
" 1
_______________________________________________________________________________________
Example 13) Leg and leg
A = 57 ° 5 9 " 4 1 " " a = 2 feet
B = 32 ° 0 " 1 9 " " b = 15 inches
C = 90 ° c = 28.30 inches
c = 24
2
2
A = tan
1
- Right Angle Trigonometry - 5 - www.mastermathmentor.com - Stu Schwartz
Multi-Step Problems :
Example 14) Consider the picture below. I want to find the length of segment AB. Suppose
" A = 25 °," B = 40 ° and BC = 12. Do the necessary work on the right to find
AB.
CD = 12 tan 40 ° = 10.
AC =
tan 25 °
AB = 21.59 " 12 = 9.
Example 15) Let’s tweak the problem slightly. I want to find the length of segment CD.
" A = 25 °," B = 40 ° and AB = 12. Note that we do not have any information about sides of either right triangle.
And yet, it is possible to solve this problem. How?
CD = ACtan 25 ° = ABtan 25 ° + BCtan 25 °
CD = BCtan 40 °
12 tan 25 ° + BCtan 25 ° = BCtan 40 °
12 tan 25 ° = BC tan 40 ° " tan 25 °
BC =
12 tan 25 °
tan 40 ° " tan 25 °
CD = 15.01tan 40 ° = 12.
Example 16) Using the same picture,
" A = 35 °," B = 68 ° and AB = 76.5, find the length of segment CD.
CD = ACtan 35 ° = ABtan 35 ° + BCtan 35 °
CD = BCtan 68 °
76.5tan 35 ° + BCtan 35 ° = BCtan 68 ° " 76.5tan 35 ° = BC tan 68 ° # tan 35 °
BC =
76.5tan 35 °
tan 68 ° # tan 35 °
= 30.18 " CD = 30.18tan 68 ° = 74.
Real-Life Applications
Guidelines for solving a triangle problem:
a) Draw a sketch of the problem situation. Don’t be afraid to make it large.
b) Look for right triangles and sketch them in.
c) Mark the known sides and angles and unknown sides and angles using variables.
d) Express the desired sides or angles in terms of trig functions with known quantities using the
variables in the sketch.
e) Solve the trig equation you generated and express the answer using correct units.
- Right Angle Trigonometry - 7 - www.mastermathmentor.com - Stu Schwartz
- A wire holding up a 40 foot telephone pole is 38 feet long. The wire attaches to the telephone pole 10.
feet below the top. What is the angle of elevation of the wire?
sin " =
- From a window in building A, I observe the top of building B across the 50 foot wide street at an angle
of elevation of
74 ° 2 5 ". I observe the base of building B at an angle of depression of
52 ° 1 8 ". Find the
height of building B.
tan 52 ° 1
a
tan 74 ° 2
b
a = 64.64 ft b = 174.28 ft
a + b = 243.97 ft
Bearing and Course: When ships or planes navigate, they need to have a simple way of explain what
direction they are traveling. One way is called bearing. The bearing will be in the form (N or S angle E or
W). A bearing is always drawn from the nearest north or south line. A heading (or course) is always drawn
from the north line in a clockwise direction. Following are ship directions, the bearing and course.
Bearing : N60°E
Course : 60 °
Bearing : S70°E
Course : 110 °
Bearing : S35°W
Course : 215 °
Bearing : N72° 1 8 W"
Course : 287 ° 4 2 "
Tip: in word problems, whenever you see or are asked for a bearing or heading, always look for the word
“from” and draw an x-y axis at that point.
Tip: if the bearing from A to B is
N65°E , the bearing from B to A will be the same angle but the opposite
direction:
S65°W.
- A jeep leaves its present location and travels along bearing
N62°W for 29 miles. How far north and
west of its original position is it?
sin 28 ° =
n
cos 28 ° =
w
n = 13.61 miles w = 25.61 miles
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- An airplane leaves an airport and travels due east for 255 miles. It then heads due south for 330 miles.
From its current position, along what heading should it travel to reach the airport and how far is it?
d = 255
2
2
= 417.04 miles
tan " =
" = 37.69° heading : 360 ° $ 37.69° = 322.631°
- Two small boats leave an island at the same time. Boat A travels due North for 21 miles and Boat B
travels due west for 18 miles. How far apart are the boats and what is the bearing of boat A from boat
B? How about the bearing of boat B from boat A?
- A plane leaves an airport and travels 2 hours along heading
120 ° at 175 mph. It then turns onto heading
30 ° and travels 2.5 hours at 200 mph. How far from the airport is it and from that position, what is the
heading back to the airport?
d = 18
2
2
= 27.66 miles
tan " =
A from B : N40.60°E B from A : S40.60°W
d = 350
2
2
= 610.33miles
tan " =
Airport from B : S 64.99°W
Heading : 214.99°
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d)
A = 21.75° a = 0.977 in
B = 68.25° b = 2.45 inches
C = 90 ° c = 2.638 in
tan 21.75° =
a
sin 68.25° =
c
_________________________________________________________________________________
e)
A = 28.072° a = 8
B = 61.928° b = 15
C = 90 ° c = 17
64 + b
2
= 289 sin A =
_________________________________________________________________________________
f)
A = 46 ° 3
8 a = 9.
B = 43 ° 2
2 b = 8.
C = 90 ° c = 12.
2
2
= c
2
tan A =
_________________________________________________________________________________
g)
A = 56.044° a = 2.25 miles = 11,880 feet
B = 35.956° b = 8,000 feet
C = 90 ° c = 14,322.514 feet or 2.713 miles
2
2
= c
2
tan A =
_________________________________________________________________________________
- In the figure above find segment CD if
" A = 27 °," B = 51 ° and AB = 8.2 miles
CD = ACtan 27 ° = ABtan 27 °+ BCtan 27 °
CD = BCtan 51 °
8.2tan 27 °+ BCtan 27 ° = BCtan 51 °
8.2tan 27 ° = BC( tan 51 °" tan 27 °)
BC =
8.2tan 27 °
tan 51 °" tan 27 °
CD = 5.760tan 51 ° = 7.113 miles
- Right Angle Trigonometry - 11 - www.mastermathmentor.com - Stu Schwartz
For each word problem, draw a picture, fill in given sides and use variables for sides and angles you don’t
know, then show equations (and/or trig functions) to find out the desired information and circle your
answer(s) being sure to use proper unit.
- A surveyor wants to determine the horizontal distance that the top of a slope is from a tall house at the
bottom of the slope. He measures the distance along the slope to be 125 feet. The angle of depression
created from the top of the slope to the top and bottom of the house is
17 °. Find the horizontal distance
from the top of the slope to the house.
cos 17 ° =
a
a = 119.538 ft
- What is the angle of elevation of the sun when a 6’3” man casts a 10.5-ft shadow?
tan " =
- A street in San Francisco has a 20% grade – that is it rises 20 feet for every 100 feet horizontally. Find
the angle of elevation of the street?
- A kite is 120 feet high when 650 feet of string is out. What angle of elevation does the kite make with the
ground?
- From a balloon 2500 ft high, a command post is seen with an angle of depression of
- How far is it
from a point on the ground below the balloon to the command post?
tan " =
sin " =
tan 8 ° 4 "
a
a = 16242.761 ft
- Right Angle Trigonometry - 13 - www.mastermathmentor.com - Stu Schwartz
- An airplane travels at 165 mph for 2 hours in a direction of
S62°W from Chicago. How far south and
west is the plane from Chicago?
sin 34 ° =
s
" s = 184.534 miles south
cos 34 ° =
e
" e = 273.582 miles east
- An airplane travels at 445 mph for 4.5 hours in a direction of
124 ° from Seattle. How far south and east
is the plane from Seattle?
- A ship leaves a port at 1:00 PM traveling at 13 knots directly north. Another ship leaves the same port
at 2:00 PM traveling due east at 15 knots. At 9:00 PM, how far apart are the ships? Along what
heading will the northern ship have to travel to reach the eastern ship which is now stopped?
tan " =
Heading is 134.726°
Distance : 104
2
2
Distance :147.787 miles
- A man is orienteering (traveling through forest area with only a compass). He leaves his base and
travels due north for 2.5 miles. He then travels due west for .8 miles. He then travels due south for 4
miles. How far is he from his base and along what heading must he travel to reach it?
tan " =
Heading back to base : 28.072°
Distance : 1.
2
2
Distance :1.7 miles
sin 34 ° =
s
" s = 1 ,119.784 miles south
cos 34 ° =
e
" e = 1 ,160.148 miles east
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- A small Cessna plane leaves Naples, Florida and travels 99 miles east to Miami, Florida. It then heads
due south for 123 miles where it crashes into the ocean. If helicopter rescue is to be effected from
Naples, how far and along what heading will the helicopter have to travel?
- John and Ellen have a big fight and swear they will never talk to each other again. John leaves the
airport traveling 180 miles along heading 25
o
. Ellen leaves the airport and travels 580 miles along
heading 295
o
. If Ellen decides she was wrong and decides to catch a flight to John that can travel at 500
mph, how long will it take her to fly into his arms?
The triangle has a right angle
d = 180
2
2
= 607.289 miles
Time :
= 1.2145 hrs
Answers to Problem 1
# Problem Answer # Problem Answer
a.
sin 82 °
0.9903 m.
sin
" 1
b.
cos 77 °
0.2250 n.
cos
" 1
c.
tan13.6°
0.2419 o.
arctan 6
d.
cos 25 ° 1 2 "
0.9048 p.
arccos 1.
DMS
impossible
e.
tan 225 ° 2
1.0169 q.
arcsin 0.
DMS
f.
sin 95 ° 3 0 " "
0.9962 r.
tan
" 1
3
g.
csc 49 °
1.3250 s.
csc
" 1
h.
sec " 24 °
1.0946 t.
sec
" 1
impossible
i.
cot 0.543°
105.5139 u.
arccot 2
j.
sec 45 ° 1
1.4217 v.
arcsec 3.
DMS
k.
csc11° 5
5.2331 w.
cot
" 1
5 DMS
l.
cot 2
arccsc 2.
DMS
tan " =
S 38.830° E or 141.170°
d = 99
2
2
= 157.892 miles
