1. 7 Applications of Trigonometry and Vectors
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2. Applications of Trigonometry and Vectors7
7.1 Oblique Triangles and the Law of Sines
7.2 The Ambiguous Case of the Law of Sines
7.3 The Law of Cosines
7.4 Vectors, Operations, and the Dot Product
7.5 Applications of Vectors
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3. Oblique Triangles and the Law of Sines
7.1
Oblique Triangles and the Law of Sines
Congruency and Oblique Triangles ▪ Derivation of the Law of Sines ▪ Solving SAA and ASA Triangles (Case 1) ▪ Area of a Triangle
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4. Side-Angle-Side (SAS)
Congruence Axioms
Side-Angle-Side (SAS)
If two sides and the included angle of one triangle are equal, respectively, to two sides and the included angle of a second triangle, then the triangles are congruent.
Angle-Side-Angle (ASA)
If two angles and the included side of one triangle are equal, respectively, to two angles and the included side of a second triangle, then the triangles are congruent.
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5. Congruence Axioms Side-Side-Side (SSS)
If three sides of one triangle are equal, respectively, to three sides of a second triangle, then the triangles are congruent.
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6. Oblique Triangles
Oblique triangle A triangle that is not a right triangle
The measures of the three sides and the three angles of a triangle can be found if at least one side and any other two measures are known.
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7. Data Required for Solving Oblique Triangles
Case 1 One side and two angles are known (SAA or ASA).
Case 2 Two sides and one angle not included between the two sides are known (SSA). This case may lead to more than one triangle.
Case 3 Two sides and the angle included between the two sides are known (SAS).
Case 4 Three sides are known (SSS).
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8. Note
If three angles of a triangle are known, unique side lengths cannot be found because AAA assures only similarity, not congruence.
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9. Derivation of the Law of Sines
Start with an oblique triangle, either acute or obtuse.
Let h be the length of the perpendicular from vertex B to side AC (or its extension).
Then c is the hypotenuse of right triangle ABD, and a is the hypotenuse of right triangle BDC.
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10. Derivation of the Law of Sines
In triangle ADB,
In triangle BDC,
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11. Derivation of the Law of Sines
Since h = c sin A and h = a sin C,
Similarly, it can be shown that
and
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12. In any triangle ABC, with sides a, b, and c,
Law of Sines
In any triangle ABC, with sides a, b, and c,
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13. Solve triangle ABC if A = 32.0°, C = 81.8°, and a = 42.9 cm.
Example 1
USING THE LAW OF SINES TO SOLVE A TRIANGLE (SAA)
Solve triangle ABC if A = 32.0°, C = 81.8°, and a = 42.9 cm.
Law of sines
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14. Use the Law of Sines to find c.
Example 1
USING THE LAW OF SINES TO SOLVE A TRIANGLE (SAA) (continued)
A + B + C = 180°
C = 180° – A – B
C = 180° – 32.0° – 81.8°
= 66.2°
Use the Law of Sines to find c.
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15. First find the measure of angle B.
Example 2
USING THE LAW OF SINES IN AN APPLICATION (ASA)
Jerry wishes to measure the distance across the Big Muddy River. He determines that C = °, A = 31.10°, and b = ft. Find the distance a across the river.
First find the measure of angle B.
B = 180° – A – C = 180° – 31.10° – ° = 36.00°
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16. The distance across the river is about 305.5 feet.
Example 2
USING THE LAW OF SINES IN AN APPLICATION (ASA) (continued)
Now use the Law of Sines to find the length of side a.
The distance across the river is about feet.
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17. First, find the measures of the angles in the triangle.
Example 3
USING THE LAW OF SINES IN AN APPLICATION (ASA)
Two ranger stations are on an east-west line 110 mi apart. A forest fire is located on a bearing N 42° E from the western station at A and a bearing of N 15° E from the eastern station at B. How far is the fire from the western station?
First, find the measures of the angles in the triangle.
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18. The fire is about 234 miles from the western station.
Example 3
USING THE LAW OF SINES IN AN APPLICATION (ASA) (continued)
Now use the Law of Sines to find b.
The fire is about 234 miles from the western station.
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19. Area of a Triangle (SAS)
In any triangle ABC, the area A is given by the following formulas:
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20. Note
If the included angle measures 90°, its sine is 1, and the formula becomes the familiar
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21. Find the area of triangle ABC.
Example 4
FINDING THE AREA OF A TRIANGLE (SAS)
Find the area of triangle ABC.
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22. Before the area formula can be used, we must find either a or c.
Example 5
FINDING THE AREA OF A TRIANGLE (ASA)
Find the area of triangle ABC if A = 24°40′, b = 27.3 cm, and C = 52°40′.
Draw a diagram.
Before the area formula can be used, we must find either a or c.
B = 180° – 24°40′ – 52°40′ = 102°40′
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23. Example 5 Now find the area.
FINDING THE AREA OF A TRIANGLE (ASA) (continued)
Now find the area.
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24. Caution
Whenever possible, use given values in solving triangles or finding areas rather than values obtained in intermediate steps to avoid possible rounding errors.
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25. The Ambiguous Case of the Law of Sines
7.2
The Ambiguous Case of the Law of Sines
Description of the Ambiguous Case ▪ Solving SSA Triangles (Case 2) ▪ Analyzing Data for Possible Number of Triangles
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26. Description of the Ambiguous Case
If the lengths of two sides and the angle opposite one of them are given (Case 2, SSA), then zero, one, or two such triangles may exist.
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27. If A is acute, there are four possible outcomes.
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28. If A is obtuse, there are two possible outcomes.
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29. Applying the Law of Sines
1. For any angle θ of a triangle, 0 < sin θ ≤ 1. If sin θ = 1, then θ = 90° and the triangle is a right triangle.
2. sin θ = sin(180° – θ) (Supplementary angles have the same sine value.)
3. The smallest angle is opposite the shortest side, the largest angle is opposite the longest side, and the middle-value angle is opposite the intermediate side (assuming the triangle has sides that are all of different lengths).
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30. Solve triangle ABC if B = 55°40′, b = 8.94 m, and a = 25.1 m.
Example 1
SOLVING THE AMBIGUOUS CASE (NO SUCH TRIANGLE)
Solve triangle ABC if B = 55°40′, b = 8.94 m, and a = 25.1 m.
Law of sines (alternative form)
Since sin A > 1 is impossible, no such triangle exists.
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31. An attempt to sketch the triangle leads to this figure.
Example 1
SOLVING THE AMBIGUOUS CASE (NO SUCH TRIANGLE) (continued)
An attempt to sketch the triangle leads to this figure.
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32. Note
In the ambiguous case, we are given two sides and an angle opposite one of the sides (SSA).
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33. Solve triangle ABC if A = 55.3°, a = 22.8 ft, and b = 24.9 ft.
Example 2
SOLVING THE AMBIGUOUS CASE (TWO TRIANGLES)
Solve triangle ABC if A = 55.3°, a = 22.8 ft, and b = 24.9 ft.
There are two angles between 0° and 180° such that sin B ≈ :
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34. Solve separately for triangles
Example 2
SOLVING THE AMBIGUOUS CASE (TWO TRIANGLES) (continued)
Solve separately for triangles
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35. Example 2 SOLVING THE AMBIGUOUS CASE (TWO TRIANGLES) (continued)
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36. Example 2 SOLVING THE AMBIGUOUS CASE (TWO TRIANGLES) (continued)
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37. Number of Triangles Satisfying the Ambiguous Case (SSA)
Let sides a and b and angle A be given in triangle ABC. (The law of sines can be used to calculate the value of sin B.)
1. If applying the law of sines results in an equation having sin B > 1, then no triangle satisfies the given conditions.
2. If sin B = 1, then one triangle satisfies the given conditions and B = 90°.
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38. Number of Triangles Satisfying the Ambiguous Case (SSA)
3. If 0 < sin B < 1, then either one or two triangles satisfy the given conditions.
(a) If sin B = k, then let B1 = sin–1 k and use B1 for B in the first triangle.
(b) Let B2 = 180° – B1. If A + B2 < 180°, then a second triangle exists. In this case, use B2 for B in the second triangle.
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39. Solve triangle ABC given A = 43.5°, a = 10.7 in., and c = 7.2 in.
Example 3
SOLVING THE AMBIGUOUS CASE (ONE TRIANGLE)
Solve triangle ABC given A = 43.5°, a = 10.7 in., and c = 7.2 in.
There is another angle C with sine value :
C = 180° – 27.6° = 152.4°
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40. Since c < a, C must be less than A. So C = 152.4° is not possible.
Example 3
SOLVING THE AMBIGUOUS CASE (ONE TRIANGLE) (continued)
Since c < a, C must be less than A. So C = 152.4°
is not possible.
B = 180° – 27.6° – 43.5° = 108.9°
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41. We are given that b > a, so no such triangle exists.
Example 4
ANALYZING DATA INVOLVING AN OBTUSE ANGLE
Without using the law of sines, explain why A = 104°, a = 26.8 m, and b = 31.3 m cannot be valid for a triangle ABC.
Because A is an obtuse angle, it must be the largest angle of the triangle. Thus, a must be the longest side of the triangle.
We are given that b > a, so no such triangle exists.
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42. 7.3
The Law of Cosines
Derivation of the Law of Cosines ▪ Solving SAS and SSS Triangles (Cases 3 and 4) ▪ Heron’s Formula for the Area of a Triangle
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43. Triangle Side Length Restriction
In any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.
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44. Derivation of the Law of Cosines
Let ABC be any oblique triangle located on a coordinate system as shown.
The coordinates of A are (x, y). For angle B,
and
Thus, the coordinates of A become (c cos B, c sin B).
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45. Derivation of the Law of Cosines (continued)
The coordinates of C are (a, 0) and the length of AC is b.
Using the distance formula, we have
Square both sides and expand.
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46. In any triangle, with sides a, b, and c,
Law of Cosines
In any triangle, with sides a, b, and c,
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47. Note
If C = 90°, then cos C = 0, and the formula becomes the Pythagorean theorem.
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48. Example 1
USING THE LAW OF COSINES IN AN APPLICATION (SAS)
A surveyor wishes to find the distance between two inaccessible points A and B on opposite sides of a lake. While standing at point C, she finds that AC = 259 m, BC = 423 m, and angle ACB measures 132°40′. Find the distance AB.
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49. The distance between the two points is about 628 m.
Example 1
USING THE LAW OF COSINES IN AN APPLICATION (SAS)
Use the law of cosines because we know the lengths of two sides of the triangle and the measure of the included angle.
The distance between the two points is about 628 m.
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50. Solve triangle ABC if A = 42.3°, b = 12.9 m, and c = 15.4 m.
Example 2
USING THE LAW OF COSINES TO SOLVE A TRIANGLE (SAS)
Solve triangle ABC if A = 42.3°, b = 12.9 m, and c = 15.4 m.
B < C since it is opposite the shorter of the two sides b and c. Therefore, B cannot be obtuse.
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51. Use the law of sines to find the measure of another angle.
Example 2
USING THE LAW OF COSINES TO SOLVE A TRIANGLE (SAS) (continued)
Use the law of sines to find the measure of another angle.
≈ 10.47
Now find the measure of the third angle.
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52. Caution
If we used the law of sines to find C rather than B, we would not have known whether C is equal to 81.7° or its supplement, 98.3°.
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53. Solve triangle ABC if a = 9.47 ft, b = 15.9 ft, and c = 21.1 ft.
Example 3
USING THE LAW OF COSINES TO SOLVE A TRIANGLE (SSS)
Solve triangle ABC if a = 9.47 ft, b = 15.9 ft, and c = 21.1 ft.
Use the law of cosines to find the measure of the largest angle, C. If cos C < 0, angle C is obtuse.
Solve for cos C.
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54. Now find the measure of angle A.
Example 3
USING THE LAW OF COSINES TO SOLVE A TRIANGLE (SSS) (continued)
Use either the law of sines or the law of cosines to find the measure of angle B.
Now find the measure of angle A.
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55. Find the measure of angle B in the figure.
Example 4
DESIGNING A ROOF TRUSS (SSS)
Find the measure of angle B in the figure.
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56. Four possible cases can occur when solving an oblique triangle.
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58. Heron’s Area Formula (SSS)
If a triangle has sides of lengths a, b, and c, with semiperimeter
then the area of the triangle is
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59. Example 5
USING HERON’S FORMULA TO FIND AN AREA (SSS)
The distance “as the crow flies” from Los Angeles to New York is 2451 miles, from New York to Montreal is 331 miles, and from Montreal to Los Angeles is 2427 miles. What is the area of the triangular region having these three cities as vertices? (Ignore the curvature of Earth.)
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60. Using Heron’s formula, the area  is
Example 5
USING HERON’S FORMULA TO FIND AN AREA (SSS) (continued)
The semiperimeter s is
Using Heron’s formula, the area  is
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61. 7.4 Vectors, Operations, and the Dot Product
Basic Terminology ▪ Algebraic Interpretation of Vectors ▪ Operations with Vectors ▪ Dot Product and the Angle Between Vectors
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62. Basic Terminology
Scalar: The magnitude of a quantity. It can be represented by a real number.
A vector in the plane is a directed line segment.
Consider vector OP
O is called the initial point
P is called the terminal point
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63. Basic Terminology Magnitude: length of a vector, expressed as |OP|
Two vectors are equal if and only if they have the same magnitude and same direction.
Vectors OP and PO have the same magnitude, but opposite directions. |OP| = |PO|
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64. Basic Terminology A = B C = D A ≠ E A ≠ F
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65. Two ways to represent the sum of two vectors
The sum of two vectors is also a vector.
The vector sum A + B is called the resultant.
Two ways to represent the sum of two vectors
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66. Sum of Two Vectors
The sum of a vector v and its opposite –v has magnitude 0 and is called the zero vector.
To subtract vector B from vector A, find the vector sum A + (–B).
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67. Scalar Product of a Vector
The scalar product of a real number k and a vector u is the vector k ∙ u, with magnitude |k| times the magnitude of u.
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68. Algebraic Interpretation of Vectors
A vector with its initial point at the origin is called a position vector.
A position vector u with its endpoint at the point (a, b) is written
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69. Algebraic Interpretation of Vectors
The numbers a and b are the horizontal component and vertical component of vector u.
The positive angle between the x-axis and a position vector is the direction angle for the vector.
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70. Magnitude and Direction Angle of a Vector a, b
The magnitude (length) of a vector u = a, b is given by
The direction angle θ satisfies where a ≠ 0.
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71. Find the magnitude and direction angle for u = 3, –2.
Example 1
FINDING MAGNITUDE AND DIRECTION ANGLE
Find the magnitude and direction angle for u = 3, –2.
Magnitude:
Direction angle:
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72. Graphing calculator solution:
Example 1
FINDING MAGNITUDE AND DIRECTION ANGLE (continued)
Graphing calculator solution:
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73. Horizontal and Vertical Components
The horizontal and vertical components, respectively, of a vector u having magnitude |u| and direction angle θ are given by or
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74. Horizontal component: 18.7
Example 2
FINDING HORIZONTAL AND VERTICAL COMPONENTS
Vector w has magnitude 25.0 and direction angle 41.7°. Find the horizontal and vertical components.
Horizontal component: 18.7
Vertical component: 16.6
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75. Graphing calculator solution:
Example 2
FINDING HORIZONTAL AND VERTICAL COMPONENTS
Graphing calculator solution:
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76. Write each vector in the figure in the form a, b.
Example 3
WRITING VECTORS IN THE FORM a, b
Write each vector in the figure in the form a, b.
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77. Properties of Parallelograms
1. A parallelogram is a quadrilateral whose opposite sides are parallel.
2. The opposite sides and opposite angles of a parallelogram are equal, and adjacent angles of a parallelogram are supplementary.
3. The diagonals of a parallelogram bisect each other, but do not necessarily bisect the angles of the parallelogram.
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78. Use the law of cosines with ΔPSR or ΔPQR.
Example 4
FINDING THE MAGNITUDE OF A RESULTANT
Two forces of 15 and 22 newtons act on a point in the plane. (A newton is a unit of force that equals .225 lb.) If the angle between the forces is 100°, find the magnitude of the resultant vector.
The angles of the parallelogram adjacent to P measure 80° because the adjacent angles of a parallelogram are supplementary.
Use the law of cosines with ΔPSR or ΔPQR.
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79. The magnitude of the resultant vector is about 24 newtons.
Example 4
FINDING THE MAGNITUDE OF A RESULTANT (continued)
The magnitude of the resultant vector is about 24 newtons.
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80. For any real numbers a, b, c, d, and k,
Vector Operations
For any real numbers a, b, c, d, and k,
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81. Let u = –2, 1 and v = 4, 3. Find the following.
Example 5
PERFORMING VECTOR OPERATIONS
Let u = –2, 1 and v = 4, 3. Find the following.
(a) u + v
= –2, 1 + 4, 3
= –2 + 4, 1 + 3
= 2, 4
(b) –2u
= –2 ∙ –2, 1
= –2(–2), –2(1)
= 4, –2
(c) 4u – 3v
= 4 ∙ –2, 1 – 3 ∙ 4, 3
= –8, 4 –12, 9
= –8 – 12, 4 – 9
= –20,–5
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82. Unit Vectors A unit vector is a vector that has magnitude 1.
j = 0, 1
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83. Unit Vectors
Any vector a, b can be expressed in the form ai + bj using the unit vectors i and j.
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84. Dot Product
The dot product (or inner product) of the two vectors u = a, b and v = c, d is denoted u ∙ v, read “u dot v,” and is given by
u ∙ v = ac + bd.
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85. Example 6 Find each dot product. = 2(4) + 3(–1) = 5
FINDING DOT PRODUCTS
Find each dot product.
(a) 2, 3 ∙ 4, –1
= 2(4) + 3(–1) = 5
(b) 6, 4 ∙ –2, 3
= 6(–2) + 4(3) = 0
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86. Properties of the Dot Product
For all vectors u, v, and w and real number k,
(a) u ∙ v = v ∙ u
(b) u ∙ (v + w) = u ∙ v + u ∙ w
(c) (u + v) ∙ w = u ∙ w + v ∙ w
(d) (ku) ∙ v = k(u ∙ v) = u ∙ kv
(e) 0 ∙ u = 0
(f) u ∙ u = |u|2
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87. Geometric Interpretation of the Dot Product
If θ is the angle between the two nonzero vectors u and v, where 0° ≤ θ ≤ 180°, then or
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88. Find the angle θ between the two vectors u = 3, 4 and v = 2, 1.
Example 7
FINDING THE ANGLE BETWEEN TWO VECTORS
Find the angle θ between the two vectors u = 3, 4 and v = 2, 1.
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89. Dot Products
For angles θ between 0° and 180°, cos θ is positive, 0, or negative when θ is less than, equal to, or greater than 90°, respectively.
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90. Note
If a ∙ b = 0 for two nonzero vectors a and b, then cos θ = 0 and θ = 90°. Thus, a and b are perpendicular or orthogonal vectors.
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91. Applications of Vectors
7.5
Applications of Vectors
The Equilibrant ▪ Incline Applications ▪ Navigation Applications
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92. The Equilibrant
Sometimes it is necessary to find a vector that will counterbalance a resultant.
This opposite vector is called the equilibrant.
The equilibrant of vector u is the vector –u.
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93. Example 1
FINDING THE MAGNITUDE AND DIRECTION OF AN EQUILIBRANT
Find the magnitude of the equilibrant of forces of 48 newtons and 60 newtons acting on a point A, if the angle between the forces is 50°. Then find the angle between the equilibrant and the 48-newton force.
The equilibrant is –v.
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94. The magnitude of –v is the same as the magnitude of v.
Example 1
FINDING THE MAGNITUDE AND DIRECTION OF AN EQUILIBRANT (cont.)
The magnitude of –v is the same as the magnitude of v.
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95. Example 1
FINDING THE MAGNITUDE AND DIRECTION OF AN EQUILIBRANT (cont.)
The required angle, , can be found by subtracting the measure of angle CAB from 180°. Use the law of sines to find the measure of angle CAB.
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96. Example 1
FINDING THE MAGNITUDE AND DIRECTION OF AN EQUILIBRANT (cont.)
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97. The vertical force BA represents the force of gravity.
Example 2
FINDING A REQUIRED FORCE
Find the force required to keep a 50-lb wagon from sliding down a ramp inclined at 20° to the horizontal. (Assume there is no friction.)
The vertical force BA represents the force of gravity.
BA = BC + (–AC)
Vector BC represents the force with which the weight pushes against the ramp.
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98. Vector BF represents the force that would pull the weight up the ramp.
Example 2
FINDING A REQUIRED FORCE (cont.)
Vector BF represents the force that would pull the weight up the ramp.
Since vectors BF and AC are equal, |AC| gives the magnitude of the required force.
Vectors BF and AC are parallel, so the measure of angle EBD equals the measure of angle A.
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99. Example 2
FINDING A REQUIRED FORCE (cont.)
Since angle BDE and angle C are right angles, triangles CBA and DEB have two corresponding angles that are equal and, thus, are similar triangles.
Therefore, the measure of angle ABC equals the measure of angle E, which is 20°.
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100. From right triangle ABC,
Example 2
FINDING A REQUIRED FORCE (cont.)
From right triangle ABC,
A force of approximately 17 lb will keep the wagon from sliding down the ramp.
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101. Example 3
FINDING AN INCLINE ANGLE
A force of 16.0 lb is required to hold a 40.0 lb lawn mower on an incline. What angle does the incline make with the horizontal?
Vector BE represents the force required to hold the mower on the incline.
In right triangle ABC, the measure of angle B equals θ, the magnitude of vector BA represents the weight of the mower, and vector AC equals vector BE.
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102. The hill makes an angle of about 23.6° with the horizontal.
Example 3
FINDING AN INCLINE ANGLE (cont.)
The hill makes an angle of about 23.6° with the horizontal.
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103. Example 4
APPLYING VECTORS TO A NAVIGATION PROBLEM
A ship leaves port on a bearing of 28.0° and travels 8.20 mi. The ship then turns due east and travels 4.30 mi. How far is the ship from port? What is its bearing from port?
Vectors PA and AE represent the ship’s path. We are seeking the magnitude and bearing of PE.
Triangle PNA is a right triangle, so the measure of angle NAP = 90° − 28.0° = 62.0°.
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104. The ship is about 10.9 miles from port.
Example 4
APPLYING VECTORS TO A NAVIGATION PROBLEM (continued)
Use the law of cosines to find |PE|.
The ship is about 10.9 miles from port.
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105. Now add 28.0° to 20.4° to find that the bearing is 48.4°.
Example 4
APPLYING VECTORS TO A NAVIGATION PROBLEM (continued)
To find the bearing of the ship from port, first find the measure of angle APE. Use the law of sines.
Now add 28.0° to 20.4° to find that the bearing is 48.4°.
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106. Airspeed and Groundspeed
The airspeed of a plane is its speed relative to the air.
The groundspeed of a plane is its speed relative to the ground.
The groundspeed of a plane is represented by the vector sum of the airspeed and windspeed vectors.
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107. The groundspeed is represented by |x|.
Example 5
APPLYING VECTORS TO A NAVIGATION PROBLEM
A plane with an airspeed of 192 mph is headed on a bearing of 121°. A north wind is blowing (from north to south) at 15.9 mph. Find the groundspeed and the actual bearing of the plane.
The groundspeed is represented by |x|.
We must find angle  to determine the bearing, which will be 121° + .
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108. The plane’s groundspeed is about 201 mph.
Example 5
APPLYING VECTORS TO A NAVIGATION PROBLEM (continued)
Find |x| using the law of cosines.
The plane’s groundspeed is about 201 mph.
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109. Find  using the law of sines.
Example 5
APPLYING VECTORS TO A NAVIGATION PROBLEM (continued)
Find  using the law of sines.
The plane’s groundspeed is about 201 mph on a bearing of 121° + 4° = 125°.
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