6.1 Law of Sines Objective To use Law of Sines to solve oblique triangles and to find the areas of oblique triangles.

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 An oblique triangle is a triangle that has no right angles. To solve an oblique triangle, you need to know the measure of at least one side and the measures of any other two parts of the triangle – two sides, two angles, or one angle and one side. C BA a b c

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3 The following cases are considered when solving oblique triangles. 1.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

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 The first two cases can be solved using the Law of Sines. (The last two cases can be solved using the Law of Cosines.) Law of Sines If ABC is an oblique triangle with sides a, b, and c, then Acute Triangle C BA b h c a C B A b h c a Obtuse Triangle

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5 Find the remaining angle and sides of the triangle. Example 1 (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 ft 0.83 ft

Using the Law of Sines to Solve a Triangle Example(AAS) Solve triangle ABC if A = 32.0°, B = 81.8°, and a = 42.9 centimeters. SolutionDraw the triangle and label the known values. Because A, B, and a are known, we can apply the law of sines involving these variables.

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

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8 Use the Law of Sines to solve the triangle. A = 110 , a = 125 inches, b = 100 inches Example 2 (SSA): C  180  – 110  –  C B A b = 100 in c a = 125 in 110    in = 

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9 The Ambiguous Case (SSA) Two angles and one side determine a unique triangle. But what if you are given two sides and one opposite angle? Three possible situations can occur: 1) no such triangle exists; 2) one such triangle exits; 3) two distinct triangles may satisfy the conditions.

Ambiguous Case

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 11 Use the Law of Sines to solve the triangle. A = 76 , a = 18 inches, b = 20 inches Example 3 (SSA): There is no angle whose sine is There is no triangle satisfying the given conditions. C A B b = 20 in a = 18 in 76 

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 12 Use the Law of Sines to solve the triangle. A = 58 , a = 11.4 cm, b = 12.8 cm Example 4 (SSA): 72.2  10.3 cm Two different triangles can be formed  a = 11.4 cm C A B1B1 b = 12.8 cm c 58  Example continues. C  180  – 58  – 72.2  = 49.8 

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 13 Use the Law of Sines to solve the second triangle. A = 58 , a = 11.4 cm, b = 12.8 cm Example (SSA) continued: B 2  180  – 72.2  =   C A B2B2 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 B1B1 b = 12.8 cm c 58  C  180  – 58  –  = 14.2 

Solving the Ambiguous Case: One Triangle Example Solve the triangle ABC, given A = 43.5°, a = 10.7 inches, and c = 7.2 inches. Solution The other possible value for C: C = 180° – 27.6° = 152.4°. Add this to A: 152.4° ° = 195.9° > 180° Therefore, there can be only one triangle.

Solving the Ambiguous Case: One Triangle

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 16 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 . How tall is the flagpole? Example 7 Application: 20  Flagpole height: b 70  34  16 m 14  The flagpole is approximately 9.5 meters tall. B A C

The Sine Rule A The angle of elevation of the top of a column measured from point A, is 20 o. The angle of elevation of the top of the statue is 25 o. Find the height of the statue when the measurements are taken 50 m from its base 50 m Angle BCA = 70 o Angle ACT = Angle ATC = 110 o 65 o m B T C 180 – 110 = 70 o 180 – 70 = 110 o 180 – 115 = 65 o 20 o 25 o 5o5o

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 18 Bearings A bearing measures the acute angle that a path or line of sight makes with a fixed north-south line.

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 19 Bearings N S WE N 80° W 80° N S WE N 45° E 45° N S WE 35° S 35° E

Using the Law of Sines in an Application (ASA) ExampleTwo 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? SolutionAngle BAC = 90° – 42° = 48° Angle B = 90° + 15° = 105° Angle C = 180° – 105° – 48° = 27° Using the law of sines to find b gives

c ab Area of triangle C AB ba c Must be the included angle

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 22 Area of an Oblique Triangle C BA b c a Find the area of the triangle. A = 74 , b = 103 inches, c = 58 inches Example 5: 74  103 in 58 in

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 23 Example 6 Finding the Area of a Triangular Lot Find the area of a triangular lot containing side lengths that measure 24 yards and 18 yards and form an angle of 80° A = ½(18)(24)sin80 A = yards

Assignment Pg 414 – 416; odd, 25, 27, 29 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 24