Law of Sines and Law of Cosines Digital Lesson Law of Sines and Law of Cosines

Definition: Oblique Triangles An oblique triangle is a triangle that has no right angles. C B A a b c 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. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Definition: Oblique Triangles

Solving Oblique Triangles The following cases are considered when solving oblique triangles. Two angles and any side (AAS or ASA) A C c A B c Law of sines 2. Two sides and an angle opposite one of them (SSA) C c a Law of sines 3. Three sides (SSS) a c b Law of cosines c a B Law of cosines 4. Two sides and their included angle (SAS) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Solving Oblique Triangles

Definition: Law of Sines 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 C B A b h c a C B A b h c a Acute Triangle Obtuse Triangle Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Definition: Law of Sines

Example: Law of Sines - ASA Example (ASA): Find the remaining angle and sides of the triangle. C B A b c 60 10 a = 4.5 ft The third angle in the triangle is A = 180 – C – B = 180 – 10 – 60 = 110. Use the Law of Sines to find side b and c. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Law of Sines - ASA

The Ambiguous Case (SSA) Law of Sines In earlier examples, you saw that two angles and one side determine a unique triangle. However, if two sides and one opposite angle are given, then three possible situations can occur: (1) no such triangle exists, (2) one such triangle exists, or (3) two distinct triangles satisfy the conditions.

The Ambiguous Case (SSA)

Example 3 – Single-Solution Case—SSA For the triangle in Figure 6.5, a = 22 inches, b = 12 inches, and A = 42 . Find the remaining side and angles. One solution: a  b Figure 6.5

Now you can determine that Example 3 – Solution cont’d Check for the other “potential angle” C  180  – 21.41  = 158.59 (158.59 + 42 = 200.59 which is more than 180  so there is only one triangle.) Now you can determine that C  180  – 42  – 21.41  = 116.59  Multiply each side by b Substitute for A, a, and b. B is acute.

Multiply each side by sin C. Example 3 – Solution cont’d Then the remaining side is given by Law of Sines Multiply each side by sin C. Substitute for a, A, and C. Simplify.

Example: Single Solution Case - SSA Example (SSA): Use the Law of Sines to solve the triangle. A = 110, a = 125 inches, b = 100 inches C B A b = 100 in c a = 125 in 110 C  180 – 110 – 48.74 = 21.26 Since the “given angle is already obtuse, there will only be one triangle. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Single Solution Case - SSA

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

Example: Two-Solution Case - SSA Example (SSA): a = 11.4 cm C A B1 b = 12.8 cm c 58 Use the Law of Sines to solve the triangle. A = 58, a = 11.4 cm, b = 12.8 cm C  180 – 58 – 72.2 = 49.8 Check for the other “potential angle” C  180  – 72.2  = 107.8  (107.8  + 58  = 165.8  which is less than 180  so there are two triangles.) Example continues. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Two-Solution Case - SSA

Example: Two-Solution Case – SSA continued Example (SSA) continued: 72.2 10.3 cm 49.8 a = 11.4 cm C A B1 b = 12.8 cm c 58 Use the Law of Sines to solve the second triangle. A = 58, a = 11.4 cm, b = 12.8 cm B2  180 – 72.2 = 107.8  C  180 – 58 – 107.8 = 14.2 C A B2 b = 12.8 cm c a = 11.4 cm 58 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Two-Solution Case – SSA continued

Area of an Oblique Triangle

Area of an Oblique Triangle The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle. Referring to the triangles below, that each triangle has a height of h = b sin A. A is acute. A is obtuse.

Area of a Triangle - SAS SAS – you know two sides: b, c and the angle between: A Remember area of a triangle is ½ base ● height Base = b Height = c ● sin A  Area = ½ bc(sinA) A B C c a b h Looking at this from all three sides: Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)

Area of an Oblique Triangle

Example – Finding the Area of a Triangular Lot Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102. Solution: Consider a = 90 meters, b = 52 meters, and the included angle C = 102 Then, the area of the triangle is Area = ½ ab sin C = ½ (90)(52)(sin102)  2289 square meters.

Law of Cosines (Solving for a side) (SSS and SAS) can be solved using the Law of Cosines. (Solving for a side) Always solve for the angle across from the longest side first!

Law of Cosines ( solving for an angle) Copyright © by Houghton Mifflin Company, Inc. All rights reserved.

Example: Law of Cosines - SSS Find the three angles of the triangle. C B A 8 6 12     Find the angle opposite the longest side first.   Law of Sines: Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Law of Cosines - SSS

Example: Law of Cosines - SAS B A 6.2 75 9.5 Solve the triangle. Law of Cosines: Law of Sines: Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Law of Cosines - SAS

Applications

An Application of the Law of Cosines The pitcher’s mound on a women’s softball field is 43 feet from home plate and the distance between the bases is 60 feet (The pitcher’s mound is not halfway between home plate and second base.) How far is the pitcher’s mound from first base?

Solution In triangle HPF, H = 45 (line HP bisects the right angle at H), f = 43, and p = 60. Using the Law of Cosines for this SAS case, you have h2 = f 2 + p2 – 2fp cos H = 432 + 602 – 2(43)(60) cos 45  1800.3. So, the approximate distance from the pitcher’s mound to first base is  42.43 feet.

Heron’s Formula

Heron’s Area Formula The Law of Cosines can be used to establish the following formula for the area of a triangle. This formula is called Heron’s Area Formula after the Greek mathematician Heron (c. 100 B.C.).

Area of a Triangle Law of Cosines Case - SSS B C c a b h SSS – Given all three sides Heron’s formula:

Try these Given the triangle with three sides of 6, 8, 10 find the area Given the triangle with three sides of 12, 15, 21 find the area