Albert Einstein (1879 – 1955)

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.

Find the distance of the fire from each tower.

N

PK

CS

65

N

70808030 k

m

42.4 km

15.5 km