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Transcript of Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.
Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
OBJECTIVES
The Law of Sines
Learn the vocabulary and conventions associated with solving triangles.Learn the statement and the derivation of the Law of Sines.Learn to solve AAS and ASA triangles by using the Law of Sines. Learn to solve for possible triangles in the ambiguous SSA case.
SECTION 7.1
1
2
3
4
Slide 7.1 - 3 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
LAW OF SINES
The following diagrams illustrate the Law of Sines.
Slide 7.1 - 4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
LAW OF SINES
sin A
a
sin B
b,
sin B
b
sinC
c,
sinC
c
sin A
A
sin A
a
sin B
b
sinC
c or
a
sin A
b
sin B
c
sinC
In any triangle ABC, with sides of length a, b, and c,
We can rewrite these relations in compact notation:
Slide 7.1 - 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SOLVING AAS AND ASA TRIANGLES
Step 1: Find the third angle. Find the measure of the third angle by subtracting the measures of the known angles from 180º.
Step 2: Make a chart. Make a chart of the six parts of the triangle, indicating the known parts and the parts to be computed. Make a sketch of the triangle.
Slide 7.1 - 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SOLVING AAS AND ASA TRIANGLES
Step 3: Apply the Law of Sines. Select two ratios of the Law of Sines in which three of the four quantities are known. Solve for the fourth quantity. Use the form of the Law of Sines in which the unknown quantity is in the numerator.
Step 4: Show the solution. Show the solution by completing the chart.
Slide 7.1 - 7 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2 Height of Mount Everest
From a point on a level plain at the foot of the mountain, a surveyor finds the angle of elevation of the peak of Mount Everest to be 20º. She walks 3465 meters closer (on a direct line between the first point and the base of the mountain) and finds the angle of elevation to be 23º. Estimate the height of the peak of Mount Everest to the nearest meter.
Slide 7.1 - 8 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2 Height of Mount Everest
Solution
mABC 180º 23º 157º
mC 180º 20º 157º 3º
a
sin 20º
3465
sin 3º
Slide 7.1 - 9 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solution continued
EXAMPLE 2 Height of Mount Everest
a 3465sin 20º
sin 3º22,644 meters
In triangle BCD, sin 23º h
a.
h asin 23º 22,644 sin 23º 8848 meters
Height of Mount Everest is about 8848 meters.
Slide 7.1 - 10 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SOLVING SSA TRIANGLES(AMBGUOUS CASE)
If the lengths of two sides of a triangle and the measure of the angle opposite one of theses sides are given, then depending on the measurements, there may be
1. No such triangle2. One such triangle3. Two such triangles
For this reason, Case 2 is called the ambiguous case.
Slide 7.1 - 11 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SOLVING SSA TRIANGLES(AMBGUOUS CASE)
A is an acute angle.
Slide 7.1 - 12 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SOLVING SSA TRIANGLES(AMBGUOUS CASE)
A is an acute angle.
Slide 7.1 - 13 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SOLVING SSA TRIANGLES(AMBGUOUS CASE)
A is an obtuse angle.
Slide 7.1 - 14 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SOLVING SSA TRIANGLESStep 1: Make a chart. Make a chart of the
six parts of the triangle, indicating the known parts and the parts to be computed.
Step 2: Apply the Law of Sines to the two ratios of in which three of the four quantities are known. Solve for the sine of the fourth quantity. Use the form of the Law of Sines in which the unknown quantity is in the numerator.
Slide 7.1 - 15 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SOLVING SSA TRIANGLESStep 3: If the sine of the angle, say , in
Step 2 is greater than 1, there is no triangle with the given measurements. If sin in Step 2 is between 0 and 1, go to Step 4.
Step 4: Suppose the sine of the angle in Step 2 is x, with 0 ≤ x ≤ 1. If x ≠ 1, there are two possible values for the angle:(i) = sin-1
x with 0 < < 90º and (ii) = 180º –
Slide 7.1 - 16 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SOLVING SSA TRIANGLESStep 5: If x ≠ 1, with
(known angle) + 1 < 180º and(known angle) + 2 < 180º, thenthere are two triangles. Otherwise, there is only one triangle, and if x = 1, it is a right triangle.
Step 6: Find the third angle of the triangle(s).
Step 7: Use the Law of Sines to find the remaining side(s).
Step 8: Show solution(s).
Slide 7.1 - 17 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3 Solving an SSA Triangle (No Solution)
Solve triangle ABC with A = 50º, a = 8 inches, and b = 15 inches.
SolutionStep 1 Make a chart.
Step 2 Apply the Law of Sines.
c = ?C = ?
b = 15B = ?
a = 8A = 50º
sin B
b
sin A
a
Slide 7.1 - 18 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solution continued
sin B bsin A
a
EXAMPLE 3 Solving an SSA Triangle (No Solution)
sin B 15sin 50º
8
sin B 1.44
Step 3 Since sin B ≈ 1.44 > 1 and the range of the sine function is [–1, 1], we conclude that no triangle exists that is compatible with the given information.
Slide 7.1 - 19 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 5 Solving an SSA Triangle (Two Solutions)
Solve triangle ABC with A = 47º, a = 11 meters, and c = 13 meters. Round each answer to the nearest tenth.SolutionStep 1 Make a chart.
Step 2 Apply the Law of Sines.
c = 13C = ?
b = ?B = ?
a = 11A = 47º
sinC
c
sin A
a
Slide 7.1 - 20 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solution continued
sinC csin A
a
EXAMPLE 5 Solving an SSA Triangle (Two Solutions)
sinC 13sin 47º
110.864
Step 4 Two possible values of C withsin C ≈ 0.864 areC1 ≈ sin-1
(0.864) = 59.8º andC2 ≈ 180º – 59.8º = 120.2º
Slide 7.1 - 21 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solution continued
EXAMPLE 5 Solving an SSA Triangle (Two Solutions)
Step 5 A + C1 = 47º + 59.8º = 106.8º < 180º, A + C2 = 47º + 120.2º = 167.2º < 180º.There are two triangles with the given measurements.
Step 6 ABC1 180º 47º 59.8º 73.2º , and
ABC2 180º 47º 120.2º 12.8º .
Slide 7.1 - 22 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solution continued
EXAMPLE 5 Solving an SSA Triangle (Two Solutions)
Step 7
b1 asinABC1
sin A
b1 11sin 73.2º
sin 47º
b1 14.4 meters
b2 asinABC2
sin A
b2 11sin12.8º
sin 47º
b1 3.3 meters
b1
sinABC1
a
sin A
b2
sinABC2
a
sin A
Slide 7.1 - 23 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solution continued
EXAMPLE 5 Solving an SSA Triangle (Two Solutions)
Step 8 BAC1
BAC2
c = 13C2 ≈ 120.2º
b ≈ 3.3 ≈ 12.8º
a = 11A = 47ºABC2
c = 13C1 ≈ 59.8º
b1 ≈ 14.4 ≈ 73.2º
a = 11A = 47º
ABC1
Slide 7.1 - 24 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
BEARINGSA bearing is the measure of an acute angle from due north or due south.
N 40º E N 60º W
Slide 7.1 - 25 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
BEARINGS
S 30º W S 75º E
Slide 7.1 - 26 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6 Navigating Using Bearings
A ship sailing due west at 20 miles per hour records the bearing of an oil rig at N 55.4º W. An hour and a half later the bearing of the same rig is N 66.8º E.
a. How far is the ship from the oil rig the second time?
b. How close did the ship pass to the oil rig?
Slide 7.1 - 27 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6 Navigating Using Bearings
Solutiona. In an hour and a half the ship travels (1.5)
(20) = 30 miles due west. The oil rig (C), the starting point for the ship (A), and the position of the ship after an hour and a half (B) form the vertices of the triangle ABC.
Slide 7.1 - 28 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solution continued
Then A = 90º – 55.4º = 34.6º and
The ship is approximately 20 miles from the oil rig when the second bearing is taken.
a
sin A
c
sinC
EXAMPLE 6 Navigating Using Bearings
a csin A
sinCa
30sin 34.6º
sin122.2º20 miles
B = 90º – 66.8º = 23.2º. ThusC = 180º – 34.6º – 23.2º = 122.2º.
Slide 7.1 - 29 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solution continued
b. The shortest distance between the ship and the oil rig is the length of the segment h in triangle ABC.
h asin B
h 20sin 23.2º
h 7.9 miles
sin B h
a
EXAMPLE 6 Navigating Using Bearings
The ship passes within 7.9 miles of the oil rig.