There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle. Albert Einstein (1879 – 1955) a German-born theoretical physicist who developed the theory of general relativity, effecting a revolution in physics. Chapter 6. - PowerPoint PPT Presentation
There are only two ways to live There are only two ways to live your life. your life. One is as though nothing is a One is as though nothing is a miracle. miracle. The other is as though everything The other is as though everything is a miracle. is a miracle. Albert Einstein Albert Einstein (1879 – 1955) (1879 – 1955) a German-born a German-born theoretical theoretical physicist who physicist who developed the theory developed the theory of general relativity, of general relativity,
Transcript
There are only two ways to There are only two ways to live your life. live your life.
One is as though nothing is One is as though nothing is a miracle. a miracle.
The other is as though The other is as though everything is a miracle.everything is a miracle.Albert EinsteinAlbert Einstein (1879 – 1955) (1879 – 1955) a German-born a German-born
theoretical theoretical physicist who physicist who
developed the theory developed the theory of general relativity, of general relativity, effecting a revolution effecting a revolution
in physics.in physics.
Chapter 6Chapter 6
Additional Topics inAdditional Topics in Trigonometry Trigonometry
Day I. Law of Sines Day I. Law of Sines (6.1)(6.1)
Part 1Part 1
6.1 GOAL 16.1 GOAL 1
How to use the Law of How to use the Law of Sines to solve oblique Sines to solve oblique
triangles.triangles.
Why should you learn it?Why should you learn it?
You can use the Law of You can use the Law of Sines to solve real-life Sines to solve real-life
problems, for example to problems, for example to determine the distance determine the distance
from a ranger station to a from a ranger station to a forest fireforest fire
-- a triangle that -- a triangle that has no right angleshas no right angles
Oblique TriangleOblique Triangle
you needyou need
To SolveTo Solve
2 angles and any 2 angles and any sideside
1-Today1-Today
AAV
AASS AASSAA
2 sides and an angle 2 sides and an angle opposite one of opposite one of
themthem
2-Next Time2-Next Time
AASSSS SSSSAA
3 sides3 sides
3-Later3-Later
SSSS SS
2 sides and their 2 sides and their included angleincluded angle
4-Later4-Later
SSAASS
Students know the Students know the Law of Sines.Law of Sines.
Standard 13.1Standard 13.1
Law of SinesLaw of Sines
If ABC is a triangle with If ABC is a triangle with sides a, b, and c, thensides a, b, and c, then
a b ca b csin A sin B sin Csin A sin B sin C
== ==
Law of SinesLaw of Sines
Alternate FormAlternate Form
sin A sin B sin Csin A sin B sin C a a b c b c
== ==
Students can apply Students can apply the Law of Sines to the Law of Sines to
solve problems.solve problems.
Standard 13.2Standard 13.2
Finding a Finding a MeasurementMeasurement
Example 1Example 1
Find, to the nearest Find, to the nearest meter, the distance meter, the distance across Perch Lake from across Perch Lake from point A to point B. The point A to point B. The length of AC, or b, equals length of AC, or b, equals 110 m, and measures of 110 m, and measures of the angles of the triangle the angles of the triangle are as shown.are as shown.
A
B C
c
a
b
sin B = sin B =
h
hhcc
OROR h = c sin Bh = c sin B
sin C =sin C =hhbb
OROR h = b sin Ch = b sin C
A
B C
c
a
bh
c sin Bc sin B = b sin C= b sin C c bc bsin C sin Bsin C sin B
==
A
B C40 67
110 mh
c bc bsin C sin Bsin C sin B
== 1101106767 4040
Solve.Solve.
110sin 67sin 40
=
110sin 67 = csin 40
sin 67 sin 40 c 110
c =
c 158 meters
Given Two Given Two Angles and One Angles and One
Side - AASSide - AAS
Example 2Example 2
Using the given information, solve the triangle.
AA 2525 35353.53.5
cc
bb
BB
CC
What does it mean “to What does it mean “to solve a triangle”? solve a triangle”?
Find all unknownsFind all unknowns
3.5sin 35sin 25
=
bsin 25 = 3.5sin 35
sin 25 sin 35 3.5 b
b =
b 4.8
Using the given information, solve the triangle.
AA 2525 35353.53.5
cc
4.84.8
BB
CC
C = 180 – (25 + 35)C = 120
How do we find C?
Using the given information, solve the triangle.
AA 2525 35353.53.5
cc
4.84.8
BB
CC120120
Can I use the Pythagorean Theorem to find c? Why or why not?NOT A RIGHT TRIANGLE!
3.5sin 120sin 25
=
csin 25 = 3.5sin 120
sin 25 sin 120 3.5 c
c =
c 7.2
Using the given information, solve the triangle.
AA 2525 35353.53.5
7.27.2
4.84.8
BB
CC120120
Your TurnYour Turn
Using the given information, solve the triangle.
AA
1351351010aa
4545
bb
BB
CC
45sin 10sin 135
=
bsin 135 = 45sin 10
sin 135 sin 10 45 b
b =
b 11.0
Using the given information, solve the triangle.
AA
1351351010aa
4545
11.011.0
BB
CC
A = 180 – (135 + 10)A = 35
Using the given information, solve the triangle.
AA
1351351010aa
4545
11.011.0
BB
CC
3535
45sin 35sin 135
=
asin 135 = 45sin 35
sin 135 sin 35 45 a
a =
a 36.5
Using the given information, solve the triangle.
AA
135135101036.536.5
4545
11.011.0
BB
CC
3535
Finding a Finding a MeasurementMeasurement
Example 3Example 3
A pole tilts away from A pole tilts away from the sun at an 8° angle the sun at an 8° angle from the vertical, and it from the vertical, and it casts a 22 foot shadow. casts a 22 foot shadow. The angle of elevation The angle of elevation from the tip of the from the tip of the shadow to the top of the shadow to the top of the pole is 43°. How tall is pole is 43°. How tall is the pole?the pole?
AB
C
88
434322’22’
What do we need to find in order to use AAS or ASA?
pp
CBA
AB
C
88
434322’22’
pp = 82CBA = 90 - 8
8282
BCA = 180 - (82 + 43) = 55
5555
22sin 43sin 55
=
psin 55 = 22sin 43
sin 55 sin 43 22 p
p =
p 18.3 ft
What was the What was the psychiatrist’s reply psychiatrist’s reply
when a patient when a patient exclaimed, “I’m a exclaimed, “I’m a
teepee. I’m a teepee. I’m a wigwam!”? wigwam!”?
““Relax… You’re too Relax… You’re too tense!” tense!”
Given Two Given Two Angles and One Angles and One
Side - ASASide - ASA
Example 4Example 4
Using the given information, solve the triangle.
A = 102.4, C = 16.7, and b = 21.6
BB CC
AA
cc 102.4102.4
16.716.7
21.621.6
aa
B=180 – (102.4 + 16.7) = 60.9
60.960.9
21.6sin 102.4sin 60.9
=
21.6sin 102.4 = asin 60.9
sin 102.4 sin 60.9 a 21.6
a =
a 24.1
BB CC
AA
cc 102.4102.4
16.716.7
21.621.6
24.124.160.960.9
21.6sin 16.7sin 60.9
=
21.6sin 16.7 = csin 60.9
sin 16.7 sin 60.9 c 21.6
c =
c 7.1
BB CC
AA
7.17.1 102.4102.4
16.716.7
21.621.6
24.124.160.960.9
Your TurnYour Turn
Using the given information, solve the triangle.
A = 12.4, C = 86.4, and b = 22.5
BB AA
CC
22.722.7
86.486.4
12.412.4
22.522.5 4.94.9
81.281.2
6.1 GOAL 2 6.1 GOAL 2
How to find the areas How to find the areas of oblique triangles.of oblique triangles.
Students determine Students determine the area of triangle, the area of triangle, given one angle and given one angle and
the two adjacent the two adjacent sides.sides.
Standard 14.0Standard 14.0
A
B C
c
sin B = sin B =
h
hhcc
OROR h = c sin Bh = c sin Ba
A = ½bhacsin B
Area of an Oblique Area of an Oblique TriangleTriangle
The area of any triangle is one half The area of any triangle is one half the product of the lengths of two the product of the lengths of two
sides times the sine of their sides times the sine of their included angle. That is,included angle. That is,
AreaArea = ½bc sin A= ½bc sin A
= ½ab sin C= ½ab sin C
= ½ac sin B= ½ac sin B
Finding an Area Finding an Area of a Triangleof a Triangle
Example 5Example 5
Find the area if C = Find the area if C = 848430’, a = 16, b = 20.30’, a = 16, b = 20.
Area = ½ab sin C
= ½(16)(20)sin 84.5Area ≈ 159.3 units2
Your TurnYour Turn
Find the area if A = 5Find the area if A = 515’, 15’, b = 4.5, c = 22.b = 4.5, c = 22.Area = ½bc sin A
= ½(4.5)(22)sin 515’Area ≈ 4.5 units2
Finding an AngleFinding an Angle
Example 6Example 6
Find Find C. The area is 262 C. The area is 262 ftft22, and a = 86’ and b = , and a = 86’ and b = 11’.11’.
Area = ½ab sin C
262 = ½(86)(11)sin C
sin C = 262/473
262 = 473 sin C
C ≈ 33.6
C = sin-1(262/473)
Your TurnYour Turn
Find Find B. The area is B. The area is 1492 ft1492 ft22, and a = 202’ , and a = 202’ and c = 66’.and c = 66’.
Area = ½ac sin B
1492 = ½(202)(66)sin B
sin B = 1492/6666
1492 = 6666 sin B
B ≈ 12.9
B = sin-
1(1492/6666)
ApplicationApplication
Example 7Example 7
The bearing from Pine Knob fire tower to the Colt Station fire tower is N 65 E and the two towers are 30 km apart.
A fire spotted by rangers in each tower has a bearing of N 80 E from Pine Knob and S 70 E from Colt Station.
