Find exact values of the six trigonometric functions for each angle. Rationalize denomi- nators when applicable.
12. 135° 13. -150° 14. 1020°
Find all values of u, if u is in the interval 30°, 360°2 and has the given function value.
15. sin u= 23
2 16. sec u= -22
Use a calculator to approximate the value of each expression. Give answers to six deci- mal places.
17. sin 42° 18′ 18. sec1-212° 12′2
Find a value of u in the interval 30°, 90°2 that satisfies each statement. Write each answer in decimal degrees to six decimal places.
19. tan u=2.6743210 20. csc u=2.3861147
5.4 Solutions and Applications of Right Triangles
Historical Background The beginnings of trigonometry can be traced back to antiquity. Figure 42 shows the Babylonian tablet Plimpton 322, which provides a table of secant values. The Greek mathematicians Hipparchus and Claudius Ptolemy developed a table of chords, which gives values of sines of angles between 0° and 90° in increments of 15 minutes. Until the advent of scien- tific calculators in the late 20th century, tables were used to find function values.
Applications of spherical trigonometry accompanied the study of astronomy for these ancient civilizations. Until the mid-20th century, spherical trigonometry was studied in undergraduate courses. See Figure 43.
An introduction to applications of the plane trigonometry studied in this text involves applying the ratios to sides of objects that take the shape of right triangles.
■ Historical Background
■ Significant Digits
■ Solving Triangles
■ Angles of Elevation or Depression
■ Bearing
■ Further Applications
Figure 43 Plimpton 322
Figure 42
11. Find the exact value of each variable in the figure.
36 z
x y
w 45°
30°
Figure 44 15 ft
18 ft d
Significant Digits A number that represents the result of counting, or that results from theoretical work and is not the result of measurement, is an exact number. There are 50 states in the U.S. In this statement, 50 is an exact number.
Most values obtained for trigonometric applications are measured values that are not exact. Suppose we quickly measure a room as 15 ft by 18 ft. See
Figure 44. We calculate the length of a diagonal of the room as follows.
d2 = 152 + 182 Pythagorean theorem d2 = 549 Apply the exponents and add.
d = 2549 Square root property;
Choose the positive root.
d ≈ 23.430749
Should this answer be given as the length of the diagonal of the room? Of course not. The number 23.430749 con- tains six decimal places, while the original data of 15 ft and 18 ft are accurate only to the nearest foot. The results of a calculation can be no more accurate than the least accurate number in the calculation. Thus, the diagonal of the 15-by-18-ft room is approximately 23 ft.
If a wall measured to the nearest foot is 18 ft long, this actually means that the wall has length between 17.5 ft and 18.5 ft. If the wall is measured more accu- rately as 18.3 ft long, then its length is really between 18.25 ft and 18.35 ft. The results of physical measurement are only approximately accurate and depend on the precision of the measuring instrument as well as the aptness of the observer.
The digits obtained by actual measurement are significant digits. The measure- ment 18 ft is said to have two significant digits; 18.3 ft has three significant digits.
In the following numbers, the significant digits are identified in color.
408 21.5 18.00 6.700 0.0025 0.09810 7300 Notice the following.
• 18.00 has four significant digits. The zeros in this number represent mea- sured digits accurate to the nearest hundredth.
• The number 0.0025 has only two significant digits, 2 and 5, because the zeros here are used only to locate the decimal point.
• The number 7300 causes some confusion because it is impossible to deter- mine whether the zeros are measured values. The number 7300 may have two, three, or four significant digits. When presented with this situation, we assume that the zeros are not significant, unless the context of the problem indicates otherwise.
To determine the number of significant digits for answers in applications of angle measure, use the following table.
Significant Digits for Angles
Angle Measure to Nearest Examples
Write Answer to This Number of Significant Digits
Degree 62°, 36° two
Ten minutes, or nearest tenth of a degree 52° 30′, 60.4° three Minute, or nearest hundredth of a degree 81° 48′, 71.25° four Ten seconds, or nearest thousandth
of a degree 10° 52′ 20″, 21.264° five
565
5.4 Solutions and Applications of Right Triangles
To perform calculations with measured numbers, start by identifying the number with the least number of significant digits. Round the final answer to the same number of significant digits as this number. Remember that the answer is no more accurate than the least accurate number in the calculation.
Figure 45 C A
B
a
b c
When we are solving triangles, a labeled sketch is an important aid.
Solving Triangles To solve a triangle means to find the measures of all the angles and sides of the triangle. As shown in Figure 45, we use a to represent the length of the side opposite angle A, b for the length of the side opposite angle B, and so on. In a right triangle, the letter c is reserved for the hypotenuse.
Figure 46 34° 30 c = 12.7 in.
A C
B
a
b
EXAMPLE 1 Solving a Right Triangle Given an Angle and a Side Solve right triangle ABC, if A= 34° 30′ and c= 12.7 in.
SOLUTION To solve the triangle, find the measures of the remaining sides and angles. See Figure 46. To find the value of a, use a trigonometric function involving the known values of angle A and side c. Because the sine of angle A is given by the quotient of the side opposite A and the hypotenuse, use sin A.
sin A= a
c sin A=
side opposite hypotenuse
sin 34° 30′= a
12.7 A=34° 30′, c=12.7
a= 12.7 sin 34° 30′ Multiply by 12.7 and rewrite.
a= 12.7 sin 34.5° Convert to decimal degrees.
a≈ 12.710.566406242 Use a calculator.
a≈ 7.19 in. Three significant digits
To find the value of b, we could substitute the value of a just calculated and the given value of c in the Pythagorean theorem. It is better, however, to use the information given in the problem rather than a result just calculated. If an error is made in finding a, then b also would be incorrect. And, rounding more than once may cause the result to be less accurate. To find b, use cos A.
cos A= b
c cos A=
side adjacent hypotenuse
cos 34° 30′ = b
12.7 A=34° 30′, c=12.7
b= 12.7 cos 34° 30′ Multiply by 12.7 and rewrite.
b≈10.5 in. Three significant digits
Once b is found, the Pythagorean theorem can be used to verify the results.
LOOKING AHEAD TO CALCULUS The derivatives of the parametric equations x=ƒ1t2 and y=g1t2 often represent the rate of change of physical quantities, such as velocities. When x and y are related by an equation, the derivatives are related rates because a change in one causes a related change in the other. Determining these rates in calculus often requires solving a right triangle.
NOTE In Example 1, we could have found the measure of angle B first and then used the trigonometric function values of B to find the lengths of the unknown sides. A right triangle can usually be solved in several ways.
To maintain accuracy, always use given information as much as pos- sible, and avoid rounding in intermediate steps.
All that remains to solve triangle ABC is to find the measure of angle B.
A+ B= 90° A and B are complementary angles.
34° 30′ + B= 90° A=34°30′
B= 55° 30′ 90°=89° 60′; Subtract 34° 30′.
■✔ Now Try Exercise 23.
EXAMPLE 2 Solving a Right Triangle Given Two Sides Solve right triangle ABC, if a= 29.43 cm and c= 53.58 cm.
SOLUTION We draw a sketch showing the given information, as in Figure 47. One way to begin is to find angle A using the sine function.
Choose the positive square root.
sin A= a
c sin A=
side opposite hypotenuse
sin A= 29.43
53.58 a=29.43, c=53.58 sin A≈0.5492721165 Use a calculator.
A≈sin-110.54927211652 Use the inverse sine function.
A≈33.32° Four significant digits A≈33° 19′ 33.32° =33°+0.32160′2 The measure of B is approximately
90° - 33° 19′= 56° 41′. 90°=89° 60′
We now find b using the Pythagorean theorem.
a2+ b2= c2 Pythagorean theorem 29.432+ b2= 53.582 a=29.43, c=53.58
b2 = 53.582 - 29.432 Subtract 29.432.
b= 22004.6915 Simplify on the right; square root property b≈44.77 cm
■✔ Now Try Exercise 33.
Figure 47
c = 53.58 cm
C A B
b a = 29.43 cm
567
5.4 Solutions and Applications of Right Triangles
CAUTION Be careful when interpreting the angle of depression. Both the angle of elevation and the angle of depression are measured between the line of sight and a horizontal line.
Angles of Elevation or Depression In applications of right triangles, the angle of elevation from point X to point Y (above X) is the acute angle formed by ray XY and a horizontal ray with endpoint at X. See Figure 48(a). The angle of depression from point X to point Y (below X) is the acute angle formed by ray XY and a horizontal ray with endpoint X. See Figure 48(b).
(b) Y X
Angle of depression Horizontal
Angle of elevation Horizontal X
Y
(a)
Figure 48
EXAMPLE 3 Finding a Length Given the Angle of Elevation
At a point A, 123 ft from the base of a flagpole, the angle of elevation to the top of the flagpole is 26° 40′. Find the height of the flagpole.
SOLUTION
Step 1 See Figure 49. The length of the side adjacent to A is known, and the length of the side opposite A must be found. We will call it a.
Step 2 The tangent ratio involves the given values. Write an equation.
tan A= side opposite
side adjacent Tangent ratio tan 26° 40′ = a
123 A=26° 40′; side adjacent =123 Step 3 a= 123 tan 26° 40′ Multiply by 123 and rewrite.
a≈12310.502218882 Use a calculator.
a≈61.8 ft Three significant digits
The height of the flagpole is 61.8 ft. ■✔ Now Try Exercise 47.
123 ft
a 26° 40
A
Figure 49
Solving an Applied Trigonometry Problem
Step 1 Draw a sketch, and label it with the given information. Label the quantity to be found with a variable.
Step 2 Use the sketch to write an equation relating the given quantities to the variable.
Step 3 Solve the equation, and check that the answer makes sense.
To solve applied trigonometry problems, follow the same procedure as solv- ing a triangle. Drawing a sketch and labeling it correctly in Step 1 is crucial.
EXAMPLE 4 Finding an Angle of Depression
From the top of a 210-ft cliff, David observes a lighthouse that is 430 ft off- shore. Find the angle of depression from the top of the cliff to the base of the lighthouse.
SOLUTION As shown in Figure 50, the angle of depression is measured from a horizontal line down to the base of the lighthouse. The angle of depression and angle B, in the right triangle shown, are alternate interior angles whose measures are equal. We use the tangent ratio to solve for angle B.
tan B= side opposite
side adjacent Tangent ratio tan B = 210
430 Side opposite=210; side adjacent=430 B =tan-1a210
430b Use the inverse tangent function.
B≈26° Two significant digits
■✔ Now Try Exercise 49.
Figure 50 210 ft
430 ft A
C B
Angle of depression
Expressing Bearing (Method 1)
When a single angle is given, it is understood that bearing is measured in a clockwise direction from due north.
Bearings of 32°, 164°, 229°, and 304°
Figure 51 N
32°
N
164°
N
229°
N
304°
Several sample bearings using Method 1 are shown in Figure 51.
CAUTION A correctly labeled sketch is crucial when solving applica- tions like those that follow. Some of the necessary information is often not directly stated in the problem and can be determined only from the sketch.
Bearing We now investigate problems involving bearing, a term used in navigation. Bearing refers to the direction of motion of an object, such as a ship or airplane, or the direction of a second object at a distance relative to the ship or airplane.
We introduce two methods of measuring bearing.
569
5.4 Solutions and Applications of Right Triangles
EXAMPLE 5 Solving a Problem Involving Bearing (Method 1)
Radar stations A and B are on an east-west line, 3.7 km apart. Station A detects a plane at C, on a bearing of 61°. Station B simultaneously detects the same plane, on a bearing of 331°. Find the distance from A to C.
SOLUTION Begin with a sketch showing the given information. See Figure 52. A line drawn due north is perpendicular to an east-west line, so right angles are formed at A and B. Angles CBA and CAB can be found as follows.
∠CBA= 331°- 270° =61° and ∠CAB= 90° - 61° = 29°
A right triangle is formed. The distance from A to C, denoted b in the figure, can be found using the cosine function for angle CAB.
cos 29° = b
3.7 Cosine ratio
b =3.7 cos 29° Multiply by 3.7 and rewrite.
b≈3.2 km Two significant digits
■✔ Now Try Exercise 69.
Figure 53 shows several sample bearings using this method. Either N or S always comes first, followed by an acute angle, and then E or W.
EXAMPLE 6 Solving a Problem Involving Bearing (Method 2)
A ship leaves port and sails on a bearing of N 47° E for 3.5 hr. It then turns and sails on a bearing of S 43° E for 4.0 hr. If the ship’s rate is 22 knots (nautical miles per hour), find the distance that the ship is from port.
SOLUTION Draw and label a sketch as in
Figure 54. Choose a point C on a bearing of N 47° E from port at point A. Then choose a point B on a bearing of S 43° E from point C.
Because north-south lines are parallel, angle ACD measures 47° by alternate interior angles. The measure of angle ACB is
47° +43° = 90°, making triangle ABC a right triangle.
Expressing Bearing (Method 2)
Start with a north-south line and use an acute angle to show the direction, either east or west, from this line.
Figure 53 N
42°
N 42° E S 31° E 31°
S S 40° W
40°
S
N 52°
N 52° W Figure 52
61°
N
A 29°
B C b
N
61°
331°
3.7 km
Figure 54
B D
N
PortA
47° 43°
S S
C
47°
N
c b = 77
nautical mi
a = 88 nautical mi
Use the formula relating distance, rate, and time to find the distances in
Figure 54 from A to C and from C to B.
b= 22* 3.5= 77 nautical mi
Distance=rate*time a= 22* 4.0= 88 nautical mi
Now find c, the distance from port at point A to the ship at point B.
a2+ b2 =c2 Pythagorean theorem 882 + 772 =c2 a=88, b=77
c = 2882 + 772 If a
2+b2=c2 and c70, then c= 2a2+b2. c≈120 nautical mi Two significant digits
■✔ Now Try Exercise 75.
EXAMPLE 7 Using Trigonometry to Measure a Distance
The subtense bar method is a method that surveyors use to determine a small distance d between two points P and Q. The subtense bar with length b is cen- tered at Q and situated perpendicular to the line of sight between P and Q. See
Figure 55. Angle u is measured, and then the distance d can be determined.
Figure 55
b/2 b/2
P u d Q
Further Applications
(a) Find d when u= 1° 23′ 12″ and b= 2.0000 cm.
(b) How much change would there be in the value of d if u measured 1″ larger?
SOLUTION
(a) From Figure 55, we obtain the following.
cot u 2 = d
b 2
Cotangent ratio
d = b 2 cot u
2 Multiply and rewrite.
Let b= 2 . To evaluate u2 , we change u to decimal degrees.
1° 23′ 12″ ≈1.386666667°
Then d = 2
2 cot 1.386666667°
2 ≈ 82.634110 cm.
(b) If u is 1″ larger, then u= 1° 23′ 13″ ≈ 1.386944444°.
d = 2
2 cot 1.386944444°
2 ≈ 82.617558 cm The difference is 82.634110 - 82.617558= 0.016552 cm.
■✔ Now Try Exercise 87.
Use cot u=tan u1 to evaluate.
Figure 54 (repeated) B D
N
PortA
47° 43°
S S
C
47°
N
c b = 77
nautical mi
a = 88 nautical mi
571
5.4 Solutions and Applications of Right Triangles
EXAMPLE 8 Solving a Problem Involving Angles of Elevation
Francisco needs to know the height of a tree. From a given point on the ground, he finds that the angle of elevation to the top of the tree is 36.7°. He then moves back 50 ft. From the second point, the angle of elevation to the top of the tree is 22.2°. See Figure 56. Find the height of the tree to the nearest foot.
Figure 56 22.2°
50 ft
36.7°
A C
B h
D x
ALGEBRAIC SOLUTION
Figure 56 shows two unknowns: x, the distance from the center of the trunk of the tree to the point where the first observation was made, and h, the height of the tree. See Figure 57 in the Graphing Calculator Solution. Because nothing is given about the length of the hypotenuse of either triangle ABC or triangle BCD, we use a ratio that does not involve the hypotenuse — namely, the tangent.
In triangle ABC, tan 36.7°= h
x or h= x tan 36.7°.
In triangle BCD, tan 22.2°= h
50+ x or h = 150+ x2 tan 22.2°.
Each expression equals h, so the expressions must be equal.
x tan 36.7° =150+ x2 tan 22.2°
Equate expressions for h.
x tan 36.7° =50 tan 22.2° + x tan 22.2°
Distributive property x tan 36.7°- x tan 22.2° =50 tan 22.2°
Write the x-terms on one side.
x1tan 36.7° - tan 22.2°2 =50 tan 22.2°
Factor out x.
x = 50 tan 22.2°
tan 36.7° - tan 22.2°
Divide by the coefficient of x.
We saw above that h =x tan 36.7°. Substitute for x.
h= a 50 tan 22.2°
tan 36.7° - tan 22.2°b tan 36.7°
Use a calculator.
tan 36.7° = 0.74537703 and tan 22.2° = 0.40809244 Thus,
tan 36.7° - tan 22.2° =0.74537703 -0.40809244 = 0.33728459 and h= a5010.408092442
0.33728459 b 0.74537703≈45.
To the nearest foot, the height of the tree is 45 ft.
GRAPHING CALCULATOR SOLUTION*
In Figure 57, we have superimposed
Figure 56 on coordinate axes with the origin at D. By definition, the tangent of the angle between the x-axis and the graph of a line with equation y = mx + b is the slope of the line, m. For line DB, m = tan 22.2°. Because b equals 0, the equation of line DB is
y1 = 1tan 22.2°2x.
The equation of line AB is y2 =1tan 36.7°2x+ b.
Because b∙0 here, we use the point A150, 02 and the point-slope form to find the equation.
y2 -y0= m1x- x02 Point-slope form y2 - 0= m1x- 502 x0=50, y0=0
y2= tan 36.7°1x - 502
Lines y1 and y2 are graphed in Figure 58. The y-coordinate of the point of intersec- tion of the graphs gives the length of BC, or h. Thus, h ≈45 .
Figure 57 22.2°
50 ft
36.7°
A C
B h
D x x
y
Figure 58
−10
−20 75
150
■✔ Now Try Exercise 77.
*Source: Reprinted with permission from The Mathematics Teacher, copyright 1995 by the National Council of Teachers of Mathematics. All rights reserved.
CONCEPT PREVIEW Match each equation in Column I with the appropriate right tri- angle in Column II. In each case, the goal is to find the value of x.