8.6 Law of Sines

8.6 Law of Sines:

8.6 Law of Sines Objectives: Use the Law of Sines to solve oblique triangles (AAS or ASA) Use the Law of Sines to solve oblique triangles (SSA) Find areas of oblique triangles Use the Law of Sines to model and solve real-life problems

WHY???:

WHY??? You can use the Law of Sines to solve real-life problems involving oblique triangles. For example, you can use the Law of Sines to determine the length of the shadow of the Leaning Tower of Pisa.

Definitions:

Definitions Oblique Triangle = triangles that have no right angles To solve an oblique triangles you need the following information: Two angles and any side (AAS or ASA) Two sides and an angle opposite one of them (SSA) Three sides (SSS) Two sides and their included angle (SAS)

Slide 4:

The following cases are considered when solving oblique triangles . Solving Oblique Triangles Two angles and any side (AAS or ASA) 2. Two sides and an angle opposite one of them (SSA) 3. Three sides (SSS) 4. Two sides and their included angle (SAS) A C c A B c a c b C c a c a B

Law of Sines:

Law of Sines C B A a b c h C h A B a c b

Slide 6:

Using the Law of Sines to Solve a Triangle (AAS) Example Solve triangle ABC if A = 32.0 °, B = 81.8°, and a = 42.9 centimeters. 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.

Slide 7:

Using the Law of Sines to Solve a Triangle (AAS) Cont. To find C , use the fact that there are 180 ° in a triangle. Now we can find c .

Given Two Angles and One Side AAS:

Given Two Angles and One Side AAS Use the information to find the remaining sides an angles C = 102.3°, B = 28.7°, b = 27.4 feet (Find A, c, a) A = 24.3°, C = 54.6°, c = 2.68 (Find B, b, a)

Slide 9:

Find the remaining angle and sides of the triangle. Example: Law of Sines - ASA Example (ASA): The third angle in the triangle is A = 180  – A – B = 180  – 10 – 60 = 110. C B A b c 60  10  a = 4.5 ft 110 Use the Law of Sines to find side b and c. 4.15 ft 0.83 ft

Given Two Angles and One Side ASA:

Given Two Angles and One Side ASA Use the information to find the remaining sides an angles A = 43°, B = 98°, c = 22 ft (Find C, a, b) B = 28°, C = 104°, a = 3⅝ (Find A, b, c)

Slide 11:

Use the Law of Sines to solve the triangle. A = 110  , a = 125 inches, b = 100 inches Example: Single Solution Case - SSA Example (SSA): C  180  – 110 – 48.74 C B A b = 100 in c a = 125 in 110  48.74  21.26  48.23 in = 21.26 

Single-Solution Case -- SSA:

Single-Solution Case -- SSA Use the information to find the remaining sides an angles A = 98°, a = 10, b = 3 A = 42°, a = 22 inches, b = 12 inches

Slide 13:

Use the Law of Sines to solve the triangle. A = 76  , a = 18 inches, b = 20 inches Example: No-Solution Case - SSA Example (SSA): There is no angle whose sine is 1.078. There is no triangle satisfying the given conditions. C A B b = 20 in a = 18 in 76 

No-Solution Case -- SSA:

No-Solution Case -- SSA Use the information to find the remaining sides an angles A = 62°, a = 10, b = 12 A = 54°, a = 7, b = 10

Slide 15:

Use the Law of Sines to solve the triangle. A = 58  , a = 11.4 cm, b = 12.8 cm Example: Two-Solution Case - SSA Example (SSA): 72.2  10.3 cm Two different triangles can be formed. 49.8  a = 11.4 cm C A B 1 b = 12.8 cm c 58  Example continues. C  180  – 58 – 72.2 = 49.8 

Slide 16:

Use the Law of Sines to solve the second triangle. A = 58  , a = 11.4 cm, b = 12.8 cm Example: Two-Solution Case – SSA continued Example (SSA) continued: B 2  180  – 72.2  = 107.8  107.8  C A B 2 b = 12.8 cm c a = 11.4 cm 58  14.2  3.3 cm 72.2  10.3 cm 49.8  a = 11.4 cm C A B 1 b = 12.8 cm c 58  C  180  – 58 – 107.8 = 14.2 

Two-Solution Case -- SSA:

Two-Solution Case -- SSA Use the information to find the remaining sides an angles A = 20.5°, a = 12, b = 31

Slide 18:

18 Terminology Bearing – measure of the acute angle a path or line of sight makes with a fixed North-South line. N S 45° E W N45°E N S 70° E W S70°W

Slide 19:

Applications of Right Triangles More Bearing examples:

Slide 20:

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? 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

Slide 21:

21 Application A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14  with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20. Application: 20  Flagpole height: b 70  34 16 m 14 The flagpole is approximately 9.5 meters tall . B A C

Application #1:

Application #1 The course for a boat race starts at point A and proceeds in the direction S 52° W to point B, then in the direction S 40° E to point C, and finally back to point A. Point C lies 8 kilometers directly south of Point A. Approximate the total distance of the race course.

Application #2:

Application #2 The circular arc of a railroad curve has a chord of length 3000 feet and a central angle of 40°. Draw a figure that visually represents the problem. Show the known quantities on the figure and use variables “r” and “s” to represent the radius of the arc and the length of the arc. Find the radius “r” of the circular arc Find the length “s” of the circular arc