Download Intermediate Algebra with Trigonometry

Intermediate Algebra with Trigonometry
J. Avery
4/99
(last revised 11/03)
TOPIC
PAGE
TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES . . . . . . . . . . . . . . . . . . 2
SPECIAL TRIANGLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
FINDING TRIGONOMETRIC FUNCTIONS USING A CALCULATOR . . . . . . 9
SOLVING RIGHT TRIANGLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
TRIGONOMETRIC APPLICATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1
TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES
In right triangle ∆ABC, the sides are named using the lower-case letters corresponding to
the angles opposite the sides, as shown below. Sides a and b are called legs; c is the
hypotenuse. (Recall the Pythagorean Theorem, a2 + b2 = c2.)
A
b
C
c
a
B
Note that leg a is opposite angle A, leg b is opposite angle B and the hypotenuse c is
opposite angle C. Right triangles should always be labeled using this pattern.
The definitions of the three basic trigonometric functions based on acute angles A and B
in right ∆ABC are defined as follows:
opposite leg a
Sine A (abbreviated as “sin A”) = hypotenuse = c
adjacent leg b
Cosine A (abbreviated as “cos A”) = hypotenuse = c
opposite leg a
Tangent A (abbreviated as “tan A”) = adjacent leg = b
a
b
b
Using these basic definitions, sin B = c , cos B = c and tan B = a . Note that these are
ratios of the sides of the right triangle, not the angle degree measure.
It is often helpful to use an acronym of the first initial of the trigonometric function and
Opposite leg
their ratios to remember the definitions. Using Sin of the angle =
, Cos of
Hypotneuse
Opposite leg
Adjacent leg
and Tan of the angle =
, an acronym such as
the angle =
Hypotneuse
Adjacent leg
“Some Old Horse Caught Another Horse Taking Oats Away” can be used where the first
letter in each word stands for the trigonometric function and its ratio. Some people find
that just remembering the spelling SOHCAHTOA is helpful.
2
It is important to note that trigonometric ratios are exact values when written as
fractions. If an answer contains a radical, it should be in its simplified form. Write
answers as rounded decimals only when directed to round an answer.
Examples:
1.
Find the exact value of the three trigonometric ratios for angles A and B if a = 12,
b = 5 and c = 13.
Solution: It is helpful to sketch a right triangle and label the sides and angles.
Remember, side a is opposite angle A, side b is opposite angle B, side c is opposite angle
C.
B
opposite leg 12
sin A = hypotenuse = 13
12
adjacent leg 5
cos A = hypotenuse = 13
13
opposite leg 12
tan A = adjacent leg = 5
C
5
A
5
sin B = 13
12
cos B = 13
5
tan B = 12
2.
Find the exact value of the three trigonometric ratios for angles A and B if b = 8
and c = 10.
Solution: Use the Pythagorean Theorem to find leg a.
a2 + b2 = c2
a2 + (8)2 = (10)2
a2 + 64 = 100
a2 = 36
a=6
A
8
10
C
6 3
sin A = 10 = 5
8 4
sin B = 10 = 5
8 4
cos A = 10 = 5
6 3
cos B = 10 = 5
6 3
tan A = 8 = 4
8 4
tan B = 6 = 3
6
B
3
3.
Find the exact value of the three trigonometric ratios for angles A and B if a = 3
and b = 3. Leave answers in simplified radical form.
Solution:
A
2
2
2
a +b =c
(3)2 + (3)2 = c2
9 + 9 = c2
18 = c2
3 2=c
3 2
3
B
2 3 2
2
= 6 = 2
3 2
2
3
2
cos A =
= 2
3 2
3
tan A = 3 = 1
sin A =
4.
3
•
3
C
2
= 2
3 2
3
2
cos B =
= 2
3 2
3
tan B = 3 = 1
sin B =
3
3
If sin A = 4 , find cos A and tan A. Leave answers in simplified radical form.
opposite leg
Solution: Since sin A = hypotenuse , find the adjacent leg for the cosine and tangent
ratios.
A
a2 + b2 = c2
(3)2 + b2 = (4)2
4
9 + b2 = 16
b
b2 = 7
b= 7
C
3
B
3
3 7
7
7
Then, cos A = 4 and tan A =
•
= 7
7
7
5.
Given tan B = 4, find sin B and cos B. Leave answers in simplified radical form.
opposite leg
Solution: Since tan B = adjacent leg , and tan
the adjacent leg = 1. Using the Pythagorean
4
4 17
17 Then, sin B =
= 17 and cos B =
17
4
B = 4 = 1 , then the opposite leg = 4 and
Theorem, the hypotenuse is found to be
1
17
= 17 .
17
4
Practice Problems:
#1 - 3: Find the exact value of the three trigonometric ratios for angles A and B in right
∆ABC using the information given. Leave answers in simplified radical form.
1.
a = 1, b = 2, c = 5
2.
a = 3, c = 5
3.
a = 2, b = 5
4.
2
If cos A = 7 , find sin A and tan A.
5.
1
If tan B = 2 , find sin B and cos B.
5
SPECIAL TRIANGLES
There are two right triangles with special characteristics. One is the isosceles right
triangle. Recall that an isosceles triangle is one in which two sides are equal in length
and the angles opposite those sides are equal in measure. In an isosceles right triangle,
the two equal angles each measure 45° and the legs are equal in length. To find the ratio
of the sides in a 45°-45°-90° triangle, consider the following examples.
Examples:
In right ∆ABC, a = b = 3. The length of the hypotenuse is 3 2 (as found in the
previous example).
3
2 3 2
2
Then sin 45° =
•
= 6 = 2 ,
3 2
2
3
2 3 2
2
cos 45° =
•
= 6 = 2 ,
3 2
2
3
and
tan 45° = 3 = 1.
1.
Note that the measure of the angles in ∆ABC were known to be 45° since the legs of the
triangle are equal in length.
In ∆ABC where a and b are 4 inches long, the hypotenuse c is found to be 4 2
inches.
4
2 4 2
2
Then sin 45° =
•
= 8 = 2 ,
4 2
2
4
2 4 2
2
cos 45° =
•
= 8 = 2 ,
4 2
2
4
and
tan 45° = 4 = 1.
2.
It follows that regardless of the length of the legs in an isosceles right triangle, the three
trigonometric ratios will always reduce to
2
sin 45° = 2
A
2
cos 45° = 2
1
2
tan 45° = 1
C
1
B
The drawing at the right above shows the lengths of the legs and hypotenuse of a 45°45°-90° triangle in their lowest terms.
6
Another right triangle with special characteristics is the 30°-60°-90° triangle. The leg
opposite the 30° angle is exactly half the length of the hypotenuse. Thus in ∆ABC where
∠A = 30°, if leg a = 2, the hypotenuse c = 4; if a = 7, c = 14. If c = 10, a = 5; if c = 2,
a = 1.
Examples:
1.
Find the three trigonometric ratios of the 30° and 60° angles in right ∆ABC if c =
6
and ∠B is 30°.
Solution:
Since the leg opposite the 30° angle is half the length of the hypotenuse, b = 3. Find a by
the Pythagorean Theorem.
A
a2 + b 2 = c 2
a2 + (3)2 = (6)2
6
a2 + 9 = 36
3
2
a = 27
30°
B
C
a=3 3
a
3 1
3 3
3
sin 30° = 6 = 2
sin 60° = 6 = 2
3 3
3 1
3
cos 30° = 6 = 2
cos 60° = 6 = 2
3
3 3
3 3 3
3
tan 30° =
•
= 9 = 3
tan 60° = 3 = 3
3 3
3
Given ∠A = 30° and a = 5 cm. Find the three trigonometric functions of ∠A and
∠B in right ∆ABC.
Solution: c = 10 cm since the hypotenuse is twice the length of the leg opposite the 30°
angle. Find b using the Pythagorean Theorem.
B
a2 + b2 = c2
(5)2 + b2 = (10)2
25 + b2 = 100
5
10
2
b = 75
30°
b=5 3
C
A
b
2.
5 1
sin 30° = 10 = 2
5 3
3
cos 30° = 10 = 2
5
3 5 3
3
tan 30° =
•
= 15 = 3
5 3
3
5 3
3
sin 60° = 10 = 2
5 1
cos 60° = 10 = 2
5 3
tan 60° = 5 = 3
7
It follows that regardless of the length of the legs in a 30°-60°-90° triangle, the three
trigonometric ratios for the acute angles will always reduce to
1
sin 30° = 2
3
sin 60° = 2
1
cos 60° = 2
3
cos 30° = 2
3
tan 30° = 3
tan 60° = 3
A
30°
3
2
60°
C
1
B
The drawing above shows the lengths of the legs and hypotenuse of a 30°-60°-90°
triangle in their lowest terms.
Here is a summary table that should help you memorize the trigonometric ratios for these
special angles. Notice the pattern across the rows for sine and cosine. The tangent value
sin A ⎞
⎛
can be found by dividing sine by cosine ⎜ tan A =
cos A ⎟⎠
⎝
sin
cos
tan
30°
45°
60°
1 1
=
2 2
3
2
3
3
2
2
2
2
3
2
1 1
=
2 2
1
3
Practice Problems: Find the exact value of the trigonometric ratio. (Do not use a
calculator.)
1.
2.
3.
4.
sin 45°
tan 30°
cos 60°
tan 60°
5.
6.
7.
8.
sin 30°
cos 45°
tan 45°
sin 60°
8
FINDING TRIGONOMETRIC FUNCTIONS USING A CALCULATOR
A calculator is necessary to find the values of a trigonometric functions for angles other
than 30°, 60° and 45°. Since calculators differ in the ways input is accepted, it is
important to know how your calculator works. For some calculators, the trigonometric
function key
SIN
,
COS
, or
TAN
is pressed first, then the angle value;
for others, the angle must be entered before the trigonometric function key is pressed.
Be sure your calculator is in “degree” mode. In many calculators, the abbreviation “deg”
will appear on the display if it is in the “degree” mode. If “rad” or “grad” appears
instead, press the mode key (or the key that changes modes on your calculator) until
“deg” appears.
Try the following examples to determine how your calculator requires input of the
trigonometric function and angle.
Examples:
Answers are rounded to four decimal places.
1.
sin 25° = 0.4226
2.
cos 16° = 0.9613
3.
tan 6.45° = 0.1131
When the trigonometric ratio is known and the angle is unknown, the angle can be found
using the
SIN -1
COS -1
,
, or
TAN -1 keys. For example,
sin A = 0.879 means “find the angle A whose sine is 0.879.”
The solution is found by the equation
A = sin-1 0.879.
The calculator key sequence to find the solution is similar to the sequence used to find
the
SIN
2
5
trigonometric ratio for an angle. That is, if you used the sequence
-1
to get the solution to the first example, you will use SIN
to find the measure of angle A. If you used the sequence
use
0
.
8
7
9
SIN -1
2
0
.
8
7
5
SIN
, you will
9
to find the measure of the angle. Try the
following examples on your calculator.
9
Examples: Answers are rounded to the nearest degree.
1.
sin A = 0.879 ⇒ A = sin-1 0.879 = 62°
2.
cos A = 0.9876 ⇒ A = cos-1 0.9876 = 9°
3.
tan A = 21.0392 ⇒ A = tan-1 21.0392 = 87°
Practice Problems:
Find the trigonometric ratio of the angle rounded to four decimal places.
1.
cos 35° = _______________________________________
2.
sin 35° = _______________________________________
3.
sin 10° = _______________________________________
4.
tan 89° = _______________________________________
5.
cos 2° = _______________________________________
6.
tan 40° = _______________________________________
Find the measure of the angle to the nearest degree.
7.
cos A = 0.4512 ⇒ A = _________________________________
8.
sin A = 0.4512 ⇒ A = _________________________________
9.
tan A = 0.4512 ⇒ A = _________________________________
10.
sin A = 0.945 ⇒ A = __________________________________
11.
tan A = 10.1 ⇒ A = ___________________________________
12.
cos A = 0.3302 ⇒ A = _________________________________
10
SOLVING RIGHT TRIANGLES
Given enough information about the angles and sides of a right triangle, all unknown
sides or angles can be found. This is called “solving the right triangle.”
Examples:
1.
Given ∠A = 36° and c = 96 m in right ∆ABC, find ∠B and sides a and b, rounded
to the nearest whole meter.
Solution:
Since the sum of the angles in a triangle is 180° and ∠C is a right angle (90°),
∠B = 90° - (measure of ∠A) = 90° - 36° = 54°. The triangle in the diagram below shows
the angles and the side that are known. Use the trigonometric ratios to find a and b.
a
sin A = c
b
cos A = c
a
sin 36° = 96
b
cos 36° = 96
a = 96 (sin 36°)
a = 56 m
b = 96 (cos 36°)
b = 78 m
A
C
36°
96 m
B
2.
Given ∠B = 20° and a = 52 ft in right ∆ABC, find ∠A and sides b and c, rounded
to the nearest whole foot.
Solution:
∠A = 90° - 20° = 70°
A
b
tan B = a
a
cos B = c
b
tan 20° = 52
52
cos 20° = c
c (cos 20°) = 52
52
c=
cos 20°
c = 55 ft
b = 52 (tan 20°)
b = 19 ft
20°
C
52
B
Remember, it is often helpful to draw a right triangle and label the sides and angles that
are given. This way you should be able to quickly determine which trigonometric
functions can be used to solve for the unknown sides. Choose the functions that use the
given information, not the answers you have found so answers that are rounded will not
affect further results.
11
Practice Problems: Solve the right triangle.
1.
In right ∆ABC, ∠B = 63° and a = 5. Find ∠A and sides b and c. Round answers
to the nearest tenth.
2.
In right ∆ABC, ∠A = 15° and a = 10. Find ∠B and sides b and c. Round answers
to the nearest tenth.
12
TRIGONOMETRIC APPLICATIONS
An angle of elevation starts from the horizontal and goes upward. It is formed between
the horizontal and the line of sight when looking up at an object. Draw a horizontal ray
and a ray going upward. The angle measured between these two rays is the angle of
elevation.
angle of
elevation
horizontal
angle of
elevation
horizontal
An angle of depression starts from the horizontal and goes downward. It is formed
between the horizontal and the line of sight when looking down at an object. Draw a
horizontal ray and a ray going downward. The angle measured between these two rays is
the angle of depression.
horizontal
angle of
depression
horizontal
angle of
depression
Examples:
1.
The Washington monument is 555 feet high. If the top of the monument is 700
feet, diagonally, from a marker on the ground, what is the angle of depression
from the top of the monument to the marker? Round to the nearest degree.
Solution: Draw a sketch of the problem and identify the given parts of the right triangle.
Note that the angle of depression is equal to the angle (marked x) formed between the
ground and the diagonal. Find x using the sine function.
angle of depression
555
700
x
2.
555
sin x = 700
⎛ 555 ⎞
x = sin-1 ⎜
⎟
⎝ 700 ⎠
x = 52°
From a point on the ground, the angle of elevation to the top of a wall is 40°. If
the wall rises 8.6 meters above the ground, how far is the point on the ground
from the bottom of the wall? Round to the nearest tenth.
13
Solution:
wall
8.6 m
40°
8.6
tan 40° = x
x (tan 40°) = 8.6
8.6
x=
tan 40°
x = 10.2 m
x
3.
A loading platform is 1.2 meters above the ground. How long must a ramp be in
order to make an angle of 22° with the ground?
Solution:
platform
1.2 m
ramp
x
22°
1.2
sin 22° = x
x (sin 22°) = 1.2
1.2
x=
sin 22°
x = 3.2 m
Use the information given to determine which trigonometric function will solve the
problem. Labeling a right triangle will help identify the relationship between the known
sides and/or angles and the side or angle that is a solution to the application.
14
Practice Problems:
1.
A 25-meter high flagpole casts a shadow 10 meters long. Find the angle of
elevation of the sun to the nearest tenth of a degree.
2.
A rope is stretched from the top of a vertical pole to a point 6.8 meters from the
foot (base) of the pole. The rope makes an angle of 38° with the ground. Find the
length of the pole to the nearest tenth of a meter.
3.
From the top of a vertical cliff 400 meters high, the angle of depression to the
ground is 18°. A hang-glider is able to glide from the top of the cliff to the
ground in a straight line. What is the distance the hang-glider glides? (Round to
the nearest meter.)
15
Answers to Practice Problems
Page 5
1.
2.
3.
4.
5.
5
5
2 5
1
2 5
sin A = 5 , cos A = 5 , tan A = 2 , sin B = 5 , cos B = 5 , tan B = 2
3
4
3
4
3
4
sin A = 5 , cos A = 5 , tan A = 4 , sin B = 5 , cos B = 5 , tan B = 3
2 29
5 29
2
5 29
2 29
sin A = 29 , cos A = 29 , tan A = 5 , sin B = 29 , cos B = 29 ,
5
tan B = 2
3 5
3 5
sin A = 7 , tan A = 2
2 5
5
sin B = 5 , cos B = 5
Page 8
1.
5.
2
2
1
2
2.
6.
3
3
2
2
3.
1
2
4.
3
7.
1
8.
3
2
0.5736
0.8391
71°
3.
7.
11.
0.1736
63°
84°
4.
8.
12.
5.3 m
3.
1294 m
Page 10
1.
5.
9.
0.8192
0.9994
24°
2.
6.
10.
57.2900
27°
71°
Page 12
1.
2.
b = 9.8, c = 11
b = 37.3, c = 38.6
Page 15
1.
68.2°
2.
16