This document provides information on right triangle trigonometry including definitions of basic angle types, right triangle properties, the Pythagorean theorem, trigonometric ratios, and how to solve right triangle problems. It defines trigonometric functions like sine, cosine, and tangent in terms of an acute angle and adjacent/opposite sides. Examples are given for finding missing side lengths and converting between angle units. Practice problems apply the concepts to evaluate trig functions and solve application problems involving heights, distances, and angles of elevation/depression.

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I. Basic Facts and Definitions
1. Right angle – angle measuring °90
2. Straight angle – angle measuring °180
3. Acute angle – angle measuring between °0 and °90
4. Complementary angles – two angles whose sum is °90
5. Supplementary angles – two angles whose sum is °180
6. Right triangle – triangle with a right angle
7. Isosceles triangle – a triangle with exactly two sides equal
8. Equilateral triangle – a triangle with all three sides equal
9. The sum of the angles of a triangle is °180 .
10. In general, capital letters refer to angles while small letters refer
to the sides of a triangle. For example, side a is opposite
angle A .
II. Right Triangle Facts and Examples
1. Hypotenuse – the side opposite the right angle, sidec .
2. Pythagorean Theorem - 222
cba =+
3. A∠ and B∠ are complementary.
Right Angle
Trigonometry
B
AC
a
b
c

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III. Examples:
1. In a right triangle, the hypotenuse is 10 inches and one side is 8
inches. What is the length of the other side?
Solution: 222
cba =+
222
108 =+ b
10064 2
=+ b
362
=b
6=b
2. In a right triangle ABCΔ , if °=∠ 23A , what is the measure of B∠ ?
Solution: The two acute angles in a right triangle are
complementary.
°=∠+∠ 90BA
°=∠+° 9023 B
°=∠ 67B
IV. Similar Triangles:
a. Two triangles are similar if the angles of one triangle are equal to
the corresponding angles of the other. In similar triangles, ratios of
corresponding sides are equal.
B
AC
8=a
?=b
10=c
C
B
A
Conditions for Similar Triangles
( EGFABC ΔΔ ~ )
1. Corresponding angles in similar triangles are
equal:
EA ∠=∠
FB ∠=∠
GC ∠=∠
2. Ratios of corresponding sides are equal:
FG
BC
EF
AB
EG
AC
==
F
G E

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Example 1:
Example 2:
BC
A
F E
22
50
50
50=AE meters 22=EF m and 100=AB m
Find the length of side BC .
Notice that ABCΔ and AEFΔ are similar
since corresponding angles are equal. (There
is a right angle at both F and C , A∠ is the
same in both triangles and B∠ equals the
acute angle at E∠ .)
Solution:
BC
EF
AB
AE
= so
BC
22
100
50
=
By cross multiplying we get:
)100(22)(50 =BC
Therefore 44=BC meters.
50=AE meters 22=EF m and 100=AB m
Find the length of side BC .
a
a
a2
45
45
a
a2
a3
60
30
All °−°−° 904545 triangles are similar to one
another. Two sides are of equal length and the
hypotenuse is 2 times the length of each of the
equal sides.
All °−°−° 906030 triangles are similar to one
another. The shortest side of length a is opposite
the smallest angle ( °30 ). The hypotenuse is twice
the length of the shortest side. The side opposite
the °60 has a length 3 times the shorter leg.

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Problem: Find the lengths of the legs of a °−°−° 906030 triangle if the
hypotenuse is 8 meters.
Solution: 1) If 82 =a , then 4=a meters and 2) 34)4(33 ==a meters.
V. The Six Trigonometric Ratios for Acute Angles
TRIG TRICK: A good way to remember the trig ratios is to use the mnemonic
SOH CAH TOA!
sine =A
c
a
hypotenuse
opposite
A ==sin cosecant =A
a
c
opposite
hypotenuse
A
A ===
sin
1
csc
cosine =A
c
b
hypotenuse
adjacent
A ==cos secant =A
b
c
adjacent
hypotenuse
A
A ===
cos
1
sec
tangent =A
b
a
adjacent
opposite
A ==tan cotangent =A
a
b
opposite
adjacent
A
A ===
tan
1
cot
c
a
b
A
B
C
SOH CAH TOA
i
n
e
p
p
o
s
i
t
e
y
p
o
t
e
n
u
s
e
o
s
i
n
e
d
j
a
c
e
n
t
y
p
o
t
e
n
u
s
e
a
n
g
e
n
t
p
p
o
s
i
t
e
d
j
a
c
e
n
t

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Example 1:
Find the six trigonometric ratios for the acute angle B .
Solution:
c
b
hypotenuse
opposite
B ==sin . Using the above definitions, the rest are:
c
a
B =cos ,
a
b
B =tan ,
b
c
B =csc ,
a
c
B =sec ,
b
a
B =cot
Example 2:
In the right ABCΔ , 1=a and 3=b . Determine the six trigonometric ratios for
B∠ .
Solution:
Use Pythagorean Theorem:
222
bac +=
222
31 +=c
22
10=c
10±=c
(Since length is positive, we will only use 10=c .)
c
3=b
1=a
A
C B
10
103
10
3
sin ===
hyp
opp
B
10
10
10
1
cos ===
hyp
adj
B 3
1
3
tan ===
adj
opp
B
3
10
csc ==
opp
hyp
B 10
1
10
sec ===
adj
hyp
B
3
1
cot ==
opp
adj
B

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VI. Special Cases
a. Trigonometric values of °30 and °60 (Use the °−°−° 906030
triangle from pg. 3.)
b. Trigonometric values of °45 (Use the °−°−° 904545 triangle from
pg. 3.)
1=b
2=c
3=a
60
30 2
1
30sin =°
2
3
60sin =°
2
3
30cos =°
2
1
60cos =°
3
3
3
1
30tan ==° 3
1
3
60tan ==°
2
1
2
30csc ==°
3
32
3
2
60csc ==°
3
32
3
2
30sec ==° 2
1
2
60sec ==°
3
1
3
30cot ==°
3
3
3
1
60cot ==°
1=b
1=a
2=c
45
45
30
60
2
2
2
1
45sin ==° 2
1
2
45csc ==°
2
2
2
1
45cos ==° 2
1
2
45sec ==°
1
1
1
45tan ==° 1
1
1
45cot ==°

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VII. Converting Minutes and Seconds to Decimal Form
(Necessary for most calculator use in evaluating trig values)
1. To convert from seconds to a decimal part of a minute, divide the
number of seconds by 60.
2. To convert from minutes to a decimal part of a degree, divide the
number of minutes by 60.
Example 1: Convert 7464 ′° to degrees using decimals.
Solution:
°
⎟
⎠
⎞
⎜
⎝
⎛
+°=′°
60
47
647464
°=°+°=′° 783.64783.647464
Example 2: Convert 012115 ′′′° to degrees using decimals.
Solution: 012115012115 ′′+′°=′′′°
°
⎟
⎠
⎞
⎜
⎝
⎛
+′°=′′′°
60
10
2115012115
716.1215012115 ′°=′′′°
°
⎟
⎠
⎞
⎜
⎝
⎛
+°=′′′°
60
167.12
15012115
°=°+°=′′′° 203.15203.15012115
VIII. Right Triangle Trigonometry Problems
To Solve Right Triangle Problems:
There are six parts to any triangle; 3 sides and 3 angles. Each trig formula
(ex: sin A = a/c) contains three parts; one acute angle and two sides. If you
know values for two of the three parts then you can solve for the third
unknown part using the following method:
1. Draw a right triangle. Label the known parts with the given values and
indicate the unknown part(s) with letters.
2. To find an unknown part, choose a trig formula which involves the
unknown part and the two known parts.

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Example 1: A right triangle has 38=a and °= 61B . Find the length of side b .
Solution:
(Always check your answer by comparing size of angle and length of side; the
longer side is always opposite the larger angle.)
IX. Angles of Elevation and Depression
Example:
From a point 124 feet from the foot of a tower and on the same level, the angle of
elevation of the tower is 0236 ′° . Find the height of the tower.
Solution:
Angle of elevation
°61
b
38=a B
A
C
Which trig formulas involve an acute angle (B) and the
side opposite (b) and the side adjacent (a) to the angle?
Since both tangent and cotangent do, either could be
used to solve this problem. We will use tangent.
a
b
adj
opp
B ==tan so,
38
61tan
b
=° or
38
8040.1
b
= .
Therefore 6.68=b .
Angle of depression
h
124 ft.
0236 ′°
°=′°= )333.36tan(0236tan
24
h
)7355.0(124=h
2.91=h ft.

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Practice Problems:
1. In right triangle ABCΔ , if 39=c inches and 36=b inches, find a .
2. Find the length of side AC . Note: This problem and diagram corresponds
to finding the height of a street light pole ( AC ) if a 6 ft. man ( EF ) casts a
shadow ( BF ) of 15 ft. and the pole casts a shadow ( BC ) of 45 ft.
3. Evaluate:
a) sin E = _____________ b) tan E = _____________
c) cos F = _____________ d) sec F = _____________
4. Evaluate: (Draw reference °−°−° 906030 and °−°−° 904545 triangles)
a) sin 30 = _____________ b) tan 60 = _____________
c) sec 60 = _____________ d) tan 45 = _____________
e) csc 45 = _____________ f) cot 30 = _____________
BC
A
E
F
6
1530
F
G E
13
12
5

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5. Evaluate:
a) tan A = _____________ b) csc B = _____________
c) cot A = _____________ d) sec B = _____________
6. Convert to decimal notation using a calculator:
a) 6076 ′°
b) 317245 ′′′°
Evaluate, using a calculator:
c) 8452sin ′°
d) 2439cot ′°
7. Label the sides and remaining angles of right triangle ABCΔ , using A , B ,
a , b and c . If 43=a and °= 37A , find the values of the remaining parts.
B
C A
10
8
C

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8. Given right triangle ABCΔ with 622.0=b and 0451 ′°=A , find c . Draw a
diagram.
9. From a cliff 140 feet above the shore line, an observer notes that the angle
of depression of a ship is 0321 ′° . Find the distance from the ship to a point
on the shore directly below the observer.
cliff ship

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Answers to Right Triangle Trigonometry:
1. 15=a inches (use Pythagorean Theorem)
2.
BF
BC
EF
AC
=
15
6
45
=
AC
18=AC
3. a)
13
12
sin =E b)
5
12
tan =E c)
13
12
cos =F d)
12
13
sec =F
4. (see part E of the handout)=
a)
2
130sin =° b) 360tan =° c) 260sec =°
d) 145tan =° e) 245csc =° f) 330cot =°
5. 6=b (use Pythagorean Theorem)
a)
3
4
6
8
tan ==A b)
3
5
csc =B c)
4
3
cot =A d)
4
5
sec =B
6. a) °1.76 b) °454.45 c) 7965. d) 2045.1
7. °=∠ 53B 06.57≈b 45.71≈c
8. 1≈c (Use
c
b
A =cos or
b
c
A =sec to solve for unknown c )
9.
x
140
0321tan =′°
41.355=x ft.
C
B
A
b
a
c
cliff ship
°5.21
140
°5.21
x
(Angle of depression)