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Copyright © 2009 Pearson Addison-Wesley 7.1-2
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
7Applications of Trigonometry and Vectors
Copyright © 2009 Pearson Addison-Wesley 1.1-37.1-3
Oblique Triangles and the
Law of Sines7.1
Congruency and Oblique Triangles ▪ Derivation of the Law of
Sines ▪ Solving SAA and ASA Triangles (Case 1) ▪ Area of a
Triangle
Copyright © 2009 Pearson Addison-Wesley 1.1-47.1-4
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-57.1-5
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.
Copyright © 2009 Pearson Addison-Wesley 7.1-6
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-77.1-7
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).
Copyright © 2009 Pearson Addison-Wesley 1.1-87.1-8
Note
If three angles of a triangle are
known, unique side lengths cannot
be found because AAA assures only
similarity, not congruence.
Copyright © 2009 Pearson Addison-Wesley 7.1-9
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.
Copyright © 2009 Pearson Addison-Wesley 7.1-10
Derivation of the Law of Sines
In triangle ADB,
In triangle BDC,
Copyright © 2009 Pearson Addison-Wesley 7.1-11
Derivation of the Law of Sines
Since h = c sin A and h = a sin C,
Similarly, it can be shown that
and
Copyright © 2009 Pearson Addison-Wesley 1.1-127.1-12
Law of Sines
In any triangle ABC, with sides a, b, and c,
Copyright © 2009 Pearson Addison-Wesley 1.1-137.1-13
Example 1 USING THE LAW OF SINES TO SOLVE A
TRIANGLE (SAA)
Law of sines
Solve triangle ABC if
A = 32.0°, C = 81.8°,
and a = 42.9 cm.
Copyright © 2009 Pearson Addison-Wesley 1.1-147.1-14
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-157.1-15
Jerry wishes to measure the distance across the Big
Muddy River. He determines that C = 112.90°,
A = 31.10°, and b = 347.6 ft. Find the distance a
across the river.
Example 2 USING THE LAW OF SINES IN AN
APPLICATION (ASA)
First find the measure of angle B.
B = 180° – A – C = 180° – 31.10 – 112.90 = 36.00
Copyright © 2009 Pearson Addison-Wesley 1.1-167.1-16
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 305.5 feet.
Copyright © 2009 Pearson Addison-Wesley 1.1-177.1-17
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-187.1-18
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-197.1-19
Area of a Triangle (SAS)
In any triangle ABC, the area A is given by the following formulas:
Copyright © 2009 Pearson Addison-Wesley 1.1-207.1-20
Note
If the included angle measures 90 , its
sine is 1, and the formula becomes the
familiar
Copyright © 2009 Pearson Addison-Wesley 1.1-217.1-21
Example 4 FINDING THE AREA OF A TRIANGLE
(SAS)
Find the area of triangle ABC.
Copyright © 2009 Pearson Addison-Wesley 1.1-227.1-22
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′.
Before the area formula can be used, we must find
either a or c.
B = 180 – 24 40′ – 52 40′ = 102 40′
Draw a diagram.
Copyright © 2009 Pearson Addison-Wesley 1.1-237.1-23
Example 5 FINDING THE AREA OF A TRIANGLE
(ASA) (continued)
Now find the area.
Copyright © 2009 Pearson Addison-Wesley 1.1-247.1-24
Caution
Whenever possible, use given values in
solving triangles or finding areas rather
than values obtained in intermediate
steps to avoid possible rounding errors.
Copyright © 2009 Pearson Addison-Wesley 7.2-2
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
7Applications of Trigonometry and Vectors
Copyright © 2009 Pearson Addison-Wesley 1.1-37.2-3
The Ambiguous Case of the
Law of Sines7.2
Description of the Ambiguous Case ▪ Solving SSA Triangles
(Case 2) ▪ Analyzing Data for Possible Number of Triangles
Copyright © 2009 Pearson Addison-Wesley 7.2-4
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-77.2-7
Applying the Law of Sines
1. For any angle θ of a triangle, 0 < sin θ ≤ 1. If sin θ = 1, then θ = 90and 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).
Copyright © 2009 Pearson Addison-Wesley 1.1-87.2-8
Example 1 SOLVING THE AMBIGUOUS CASE (NO
SUCH TRIANGLE)
Law of sines(alternative form)
Solve triangle ABC if B = 55°40′, b = 8.94 m, and
a = 25.1 m.
Since sin A > 1 is impossible, no such triangle exists.
Copyright © 2009 Pearson Addison-Wesley 1.1-97.2-9
Example 1 SOLVING THE AMBIGUOUS CASE (NO
SUCH TRIANGLE) (continued)
An attempt to sketch the triangle leads to this figure.
Copyright © 2009 Pearson Addison-Wesley 1.1-107.2-10
Note
In the ambiguous case, we are given
two sides and an angle opposite one of
the sides (SSA).
Copyright © 2009 Pearson Addison-Wesley 1.1-117.2-11
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 ≈ .897867:
Copyright © 2009 Pearson Addison-Wesley 1.1-127.2-12
Example 2 SOLVING THE AMBIGUOUS CASE (TWO
TRIANGLES) (continued)
Solve separately for triangles
Copyright © 2009 Pearson Addison-Wesley 1.1-137.2-13
Example 2 SOLVING THE AMBIGUOUS CASE (TWO
TRIANGLES) (continued)
Copyright © 2009 Pearson Addison-Wesley 1.1-147.2-14
Example 2 SOLVING THE AMBIGUOUS CASE (TWO
TRIANGLES) (continued)
Copyright © 2009 Pearson Addison-Wesley 1.1-157.2-15
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 .
Copyright © 2009 Pearson Addison-Wesley 1.1-167.2-16
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-177.2-17
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 .46319186:
C = 180 – 27.6 = 152.4
Copyright © 2009 Pearson Addison-Wesley 1.1-187.2-18
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
Copyright © 2009 Pearson Addison-Wesley 1.1-197.2-19
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.
Copyright © 2009 Pearson Addison-Wesley 7.3-2
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
7Applications of Trigonometry and Vectors
Copyright © 2009 Pearson Addison-Wesley 1.1-37.3-3
The Law of Cosines7.3Derivation of the Law of Cosines ▪ Solving SAS and SSS
Triangles (Cases 3 and 4) ▪ Heron’s Formula for the Area of a
Triangle
Copyright © 2009 Pearson Addison-Wesley 1.1-47.3-4
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.
Copyright © 2009 Pearson Addison-Wesley 7.3-5
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).
Copyright © 2009 Pearson Addison-Wesley 7.3-6
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-77.3-7
Law of Cosines
In any triangle, with sides a, b, and c,
Copyright © 2009 Pearson Addison-Wesley 1.1-87.3-8
Note
If C = 90 , then cos C = 0, and the
formula becomes the
Pythagorean theorem.
Copyright © 2009 Pearson Addison-Wesley 1.1-97.3-9
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-107.3-10
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-117.3-11
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-127.3-12
Example 2 USING THE LAW OF COSINES TO
SOLVE A TRIANGLE (SAS) (continued)
≈ 10.47
Use the law of sines to find the
measure of another angle.
Now find the measure of the third angle.
Copyright © 2009 Pearson Addison-Wesley 1.1-137.3-13
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 .
Copyright © 2009 Pearson Addison-Wesley 1.1-147.3-14
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-157.3-15
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-167.3-16
Example 4 DESIGNING A ROOF TRUSS (SSS)
Find the measure of angle B in
the figure.
Copyright © 2009 Pearson Addison-Wesley 1.1-177.3-17
Four possible cases can occur when solving an
oblique triangle.
Copyright © 2009 Pearson Addison-Wesley 1.1-197.3-19
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
Copyright © 2009 Pearson Addison-Wesley 1.1-207.3-20
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.)
Copyright © 2009 Pearson Addison-Wesley 1.1-217.3-21
Example 5 USING HERON’S FORMULA TO FIND
AN AREA (SSS) (continued)
The semiperimeter s is
Using Heron’s formula, the area is
Copyright © 2009 Pearson Addison-Wesley 7.4-2
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
7Applications of Trigonometry and Vectors
Copyright © 2009 Pearson Addison-Wesley 1.1-37.4-3
Vectors, Operations, and the
Dot Product7.4
Basic Terminology ▪ Algebraic Interpretation of Vectors ▪
Operations with Vectors ▪ Dot Product and the Angle Between
Vectors
Copyright © 2009 Pearson Addison-Wesley 7.4-4
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
Copyright © 2009 Pearson Addison-Wesley 7.4-5
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 POhave the same magnitude, but opposite directions.|OP| = |PO|
Copyright © 2009 Pearson Addison-Wesley 7.4-7
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
Copyright © 2009 Pearson Addison-Wesley 7.4-8
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).
Copyright © 2009 Pearson Addison-Wesley 7.4-9
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.
Copyright © 2009 Pearson Addison-Wesley 7.4-10
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
Copyright © 2009 Pearson Addison-Wesley 7.4-11
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-127.4-12
Magnitude and Direction Angle
of a Vector a, b
The magnitude (length) of a vector u = a, bis given by
The direction angle θ satisfieswhere a ≠ 0.
Copyright © 2009 Pearson Addison-Wesley 1.1-137.4-13
Example 1 FINDING MAGNITUDE AND DIRECTION
ANGLE
Find the magnitude and direction angle for u = 3, –2 .
Magnitude:
Direction angle:
Copyright © 2009 Pearson Addison-Wesley 1.1-147.4-14
Example 1 FINDING MAGNITUDE AND DIRECTION
ANGLE (continued)
Graphing calculator solution:
Copyright © 2009 Pearson Addison-Wesley 1.1-157.4-15
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
Copyright © 2009 Pearson Addison-Wesley 1.1-167.4-16
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
Copyright © 2009 Pearson Addison-Wesley 1.1-177.4-17
Example 2 FINDING HORIZONTAL AND VERTICAL
COMPONENTS
Graphing calculator solution:
Copyright © 2009 Pearson Addison-Wesley 1.1-187.4-18
Example 3 WRITING VECTORS IN THE FORM a, b
Write each vector in the figure in
the form a, b .
Copyright © 2009 Pearson Addison-Wesley 1.1-197.4-19
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-207.4-20
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-217.4-21
Example 4 FINDING THE MAGNITUDE OF A
RESULTANT (continued)
The magnitude of the resultant vector is about24 newtons.
Copyright © 2009 Pearson Addison-Wesley 1.1-227.4-22
Vector Operations
For any real numbers a, b, c, d, and k,
Copyright © 2009 Pearson Addison-Wesley 1.1-237.4-23
Example 5 PERFORMING VECTOR OPERATIONS
Let u = –2, 1 and v = 4, 3 . Find the following.
(a) u + v
(b) –2u
(c) 4u – 3v
= –2, 1 + 4, 3 = –2 + 4, 1 + 3 = 2, 4
= –2 ∙ –2, 1 = –2(–2), –2(1) = 4, –2
= 4 ∙ –2, 1 – 3 ∙ 4, 3
= –8, 4 – 12, 9
= –8 – 12, 4 – 9 = –20,–5
Copyright © 2009 Pearson Addison-Wesley 7.4-24
Unit Vectors
A unit vector is a vector that has magnitude 1.
i = 1, 0 j = 0, 1
Copyright © 2009 Pearson Addison-Wesley 7.4-25
Unit Vectors
Any vector a, b can be expressed in the form
ai + bj using the unit vectors i and j.
Copyright © 2009 Pearson Addison-Wesley 1.1-267.4-26
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-277.4-27
Example 6 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
Copyright © 2009 Pearson Addison-Wesley 1.1-287.4-28
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
Copyright © 2009 Pearson Addison-Wesley 1.1-297.4-29
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
Copyright © 2009 Pearson Addison-Wesley 1.1-307.4-30
Example 7 FINDING THE ANGLE BETWEEN TWO
VECTORS
Find the angle θ between the two vectors u = 3, 4
and v = 2, 1 .
Copyright © 2009 Pearson Addison-Wesley 7.4-31
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-327.4-32
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.
Copyright © 2009 Pearson Addison-Wesley 7.5-2
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
7Applications of Trigonometry and Vectors
Copyright © 2009 Pearson Addison-Wesley 1.1-37.5-3
Applications of Vectors7.5The Equilibrant ▪ Incline Applications ▪ Navigation Applications
Copyright © 2009 Pearson Addison-Wesley 7.5-4
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-57.5-5
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-67.5-6
Example 1 FINDING THE MAGNITUDE AND
DIRECTION OF AN EQUILIBRANT (cont.)
The magnitude of –v is the same as the magnitude of v.
Copyright © 2009 Pearson Addison-Wesley 1.1-77.5-7
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-87.5-8
Example 1 FINDING THE MAGNITUDE AND
DIRECTION OF AN EQUILIBRANT (cont.)
Copyright © 2009 Pearson Addison-Wesley 1.1-97.5-9
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-107.5-10
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-117.5-11
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°.
Copyright © 2009 Pearson Addison-Wesley 1.1-127.5-12
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-137.5-13
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-147.5-14
Example 3 FINDING AN INCLINE ANGLE (cont.)
The hill makes an angle of about 23.6 with the
horizontal.
Copyright © 2009 Pearson Addison-Wesley 1.1-157.5-15
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 .
Copyright © 2009 Pearson Addison-Wesley 1.1-167.5-16
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-177.5-17
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 .
Copyright © 2009 Pearson Addison-Wesley 7.5-18
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.
Copyright © 2009 Pearson Addison-Wesley 1.1-197.5-19
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 + .
Copyright © 2009 Pearson Addison-Wesley 1.1-207.5-20
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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