Objectives:1. Be able to prove the Law of Sines using

Right Triangle Trigonometry.2. Be able to apply the Law of Sines on various

triangles.3. Be able to determine the number of

triangles that exist in the ambiguous case (SSA).

Critical Vocabulary:Soh-Cah-Toa, Sine Function, Ambiguous Case

Review ProblemFind: Height

a

xTan 17

17

17aTanx

feeta 83.243

a

xTan

30053

53)300( Tanax

feetx 55.74

5353)300(17 TanaaTan

535330017 aTanTanaTan

533005317 TanaTanaTan

53300)5317( TanTanTana

5317

53tan300

TanTana

17

x

a

y

53

x

300-a

z

c

Siny

b

Sin

a

Sin

The Law of Sines is three different formulas that help solve triangles that are not right triangles. These formulas do also work on right triangles

The formula states the ratio of the sine of an angle and its opposite side is equal to sine of another angle and its opposite side.

Take a look at the first of three formulas using angle alpha and angle beta.

Take a look at the second of three formulas using angle beta and angle gamma.

Take a look at the last formula using angle alpha and angle gamma.

Where did these formulas actually come from? On the next slide we will see where these formulas came from.

These formulas will help us solve: SAA, SSA, and ASA triangles.

Let’s drop down an ALTITUDE from one of the

vertices

An ALTITUDE will form 2 right angles. However, it does not divide the length into two equal parts.

Lets label the altitude of the triangle.x

a

b

c Let’s look at the right triangle on the left. Using the alpha angle find the sine (opposite/hypotenuse)

c

xSin

xcSin

Let’s look at the right triangle on the right. Using the gamma angle find the sine (opposite/hypotenuse)

a

xSin

xaSin

cSinaSin

a

cSinSin

a

Sin

c

Sin

Set the two equations equal since they both equal “x”

Divide both sides by “a” Divide both sides by “c” Now, Prove the other 2/3 of the formula. Redraw the triangle and move the altitude to another vertice.

You have proven 1/3 of the formula.

Example1:Solve the triangle that has α = 40 degrees, β = 60

degrees, γ = 80 degrees, and side length a is 4 meters.

40

4 meters

60

b

80

c

b

SinSin 60

4

40

60440 SinbSin

40

604

Sin

Sinb

metersb 39.5c

SinSin 80

4

40

80440 SincSin

40

804

Sin

Sinc metersc 13.6

Since we know an angle and its opposite side we can use the Law of Sines.

If you set up the law of sines and get a calculator error, then there is no triangle.

If you find the missing angle, and the sum of its supplement and the given angle are less than 180 degrees then there are 2 different triangles.

If you find the missing angle, and the sum of its supplement and the given angle are greater than 180 degrees then there is 1 triangle.

Example 1: a = 3 b =2, and alpha = 40 degrees

Answer: beta = 25.4 degrees

Supplement of beta is 154.6 degrees

The supplement (154.6) and the given angle (40) has a sum that is greater than 180 degrees, Therefore there is one triangle.

40

3

β

2

γ

cb

Sin

a

Sin

23

40 SinSin

SinSin 3402

3

402SinSin

Example 2: a = 6 b = 8, and alpha = 35 degrees

Answer: beta = 49.9 degrees

Supplement of beta is 130.1 degrees

The supplement (130.1) and the given angle (35) has a sum that is less than 180 degrees, Therefore there is two triangles.

6

835

β

γ

cb

Sin

a

Sin

86

35 SinSin

SinSin 6358

6

358SinSin

Example 2: a = 6 b = 8, and alpha = 35 degrees

Since beta could be 49.9 degrees or 130.1 degrees, we have two possible triangles using the given information. Now you can solve the triangles to obtain the values for c and gamma.

6

835

49.9

γ

c

358

c 6

γ

130.1

2

b

α

β

50

1c

Sin

a

Sin

1

50

2

SinSin

SinSin 502

Example 3: a = 2 c = 1, and gamma = 50 degrees

Answer: Alpha = Error

Since there was a calculator error, there is no triangle.

WARM UP Using the Law of SinesFind: Height

17

feetx 54.74

53

y

y

SinSin 17

300

110

110

17300

Sin

Siny

feety 34.93

34.9353

xSin

xSin 5334.93

Packet: Page 502-506 #9,13,21,25,29,31,37-44 all,47,51,57