Copyright © 2007 Pearson Education, Inc. Slide 10-1 While you wait: Without consulting any...

Copyright © 2007 Pearson Education, Inc. Slide 10-1

While you wait:

• Without consulting any resources or asking your friends… write down everthing you remember about the:

The Law of SinesTHE LAW OF COSINES


Sec 9.3 The Law of Sines

Oblique Triangles


Recall• In a triangle, the sum of the interior

angles is 180º.

• No triangle can have two obtuse angles.

• The height of a triangle is less than or equal to the length of two of the sides.

.


Recall

The sine function has a range of 1sin1

• If the θ is a positive decimal < 1, the θ can lie in the first quadrant (acute <) or in the second quadrant (obtuse <).

)180sin(sin • .


9.3 Data Required for Solving Oblique Triangles

Case 1 One side and two angles known: »SAA or ASA

Case 2 Two sides and one angle not included between the sides known:

»SSA»This case may lead to more than one

solution.

Case 3 Two sides and one angle included between the sides known:

»SASCase 4 Three sides are known:

»SSS


9.3 Derivation of the Law of Sines

For our purposes we will usethe triangles labelled as :

The orientation of the triangle will not make a difference.


9.3 Derivation of the Law of Sines

Start with an acute or obtuse triangle and construct the perpendicular from B to side AC. Let h be the height of this perpendicular. Then c and a are thehypotenuses of right triangle ADB and BDC, respectively.

,sinorsin AchchA

CahahC sinorsin

Cc

AaAcCa

sinsinsinsin


9.3 The Law of Sines

In a similar way, by constructing perpendiculars from other vertices, the following theorem can be proven.

The Law of Sine

Use to find ANGLES Use to find SIDES


Yet another version of the Law of Sines

Can you write two other versions?


9.3 Using the Law of Sines to Solve a Triangle

Example Solve triangle ABC if A = 32.0°, B = 81.8°,

and a = 42.9 centimeters.

B

b

A

a

sinsin

Solution Draw the triangle and label the known values.Because A, B, and a are known,we can apply the law of sines involving these variables.

8.81sin0.32sin

9.42 b cm 1.80 b


9.3 Using the Law of Sines to Solve a Triangle

To find C, use the fact that there are 180° in a triangle.180 CBA

C

c

A

a

sinsin

CAC 180 2.668.810.32180

Now we can find c

cm 1.742.66sin0.32sin

9.42 c

c


Law of Sines Day 2For each given question 1-9:• Draw a triangle with proper labels and

place the given values on the diagram.• Identify the combination given i.e. SAS,

etc.

We will be using these diagrams to solve each triangle and to record notes… so allow room around each shape for further writing.


Example𝐴=28° ;𝑎=5 ,𝑏=8

5

8𝟐𝟖°

AASor

SAA


For each diagram:

• Identify whether L.O.S. can be used.• If LOS can be used: How many

solutions:0, 1 or 2?• Can you make some generalization; or write

a program?• As you make any generalization, update

your triangle table.


One more clue:If angles and sides are ordered numerically, thenIn every triangle the smallest angle faces the smallest side

And In every triangle the middle angle faces the middle side

And

In every triangle the largest angle faces the largest side


Repeat the process with the following


9.3 Using the Law of Sines in an Application (ASA)

Example Two stations are on an east-west line 110 miles apart. A forest fire is located on a bearing of 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?

miles. 234

27sin110

105sin

b

b

Solution Angle BAC = 90° – 42° = 48°Angle B = 90° + 15° = 105°Angle C = 180° – 105° – 48° = 27°

Using the law of sines to find b gives


9.3 Ambiguous Case

• If given the lengths of two sides and the angle opposite one of them, it is possible that 0, 1, or 2 such triangles exist.

• Some basic facts that should be kept in mind:– For any angle , –1 sin 1, if sin = 1, then

= 90° and the triangle is a right triangle.– sin = sin(180° – ).– 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 unequal sides).


9.3 Number of Triangles Satisfying the Ambiguous Case

Let sides a and b and angle A be given in triangle ABC. (The law of sines can be used to calculate sin B.)

1. If 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°.

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.


9.3 Ambiguous Case

a < b sinA

a = b sinA

a = b sinA

a < b sinA

a > b sinA


How does it work?

• Web • demo

http://jwilson.coe.uga.edu/EMT668/EMAT6680.2001/Mealor/EMAT%206700/law%20of%20sines/Law%20of%20Sines%20ambiguous%20case/lawofsinesambiguouscase.html


9.3 Ambiguous Case for Obtuse Angle A


9.3 Solving the Ambiguous Case: One Triangle

Example Solve the triangle ABC, given A = 43.5°,

a = 10.7 inches, and c = 7.2 inches.

7.10

5.43sin

2.7

sin

C

Solution

46319186.sin C 6.27 C

The other possible value for C: C = 180° – 27.6° = 152.4°.

Add this to A: 152.4° + 43.5° = 195.9° > 180°

Therefore, there can be only one triangle.


9.3 Solving the Ambiguous Case: One Triangle

9.1085.436.27180 B

inches 7.145.43sin

7.109.108sin

b

b


9.3 Solving the Ambiguous Case:No Such Triangle

Example Solve the triangle ABC if B = 55°40´, b = 8.94 meters, and a = 25.1 meters.

Solution Use the law of sines to find A.

b

B

a

A sinsin

94.8

'4055sin

1.25

sin

A

3184379.2sin A

Since sin A cannot be greater than 1, the triangle does not exist.


Example

Solve the triangle ABC if A = 55.3°, a = 22.8 feet, and b = 24.9 feet.

Solution

9.3 Solving the Ambiguous Case: Two Triangles

b

B

a

A sinsin

9.24

sin

8.22

3.55sin B

8978678.sin B 9.631 B 1.1169.631802 B



To see if B2 = 116.1° is a valid possibility, add 116.1° to the measure of A: 116.1° + 55.3° = 171.4°. Since this sum is less than 180°, it is a valid triangle.

Now separate the triangles into two: AB1C1 and AB2C2. 8.609.633.55180180 11 BAC

1

1

sinsin C

c

A

a

8.60sin3.55sin

8.22 1c

feet 2.241 c



Now solve for triangle AB2C2.

6.81.1163.55180180 22 BAC

2

2

sinsin C

c

A

a

6.8sin3.55sin

8.22 2c

feet 15.42 c


Practice:

Answer in pairs.

Find m B, m C, and c, if they exist.

1) a = 9.1, b = 12, mA = 35o

2) a = 25, b = 46, mA = 37o

3) a = 15, b = 10, mA = 66o


Answers:

1)Case 1:

mB=49o,mC=96o,c=15.78

Case 2:

mB=131o,mC=14o,c=3.84

2)No possible solution.

3)mB=38o,mC=76o,c=15.93


Practice Problems on your own:

• The problems below is a sampling of what you should try; feel free to try other problems also:

• Sec 9-3; Written Exercises,• page 347 #8,14, 17-22 all

Copyright © 2007 Pearson Education, Inc. Slide 10-1 While you wait: Without consulting any...

Documents

Transcript of Copyright © 2007 Pearson Education, Inc. Slide 10-1 While you wait: Without consulting any...