7.2 LAW OF COSINES

What You Should Learn Use the Law of Cosines to solve oblique triangles (SSS or SAS). Use the Law of Cosines to model and solve real-life problems.

Introduction Two cases remain in the list of conditions needed to solve an oblique triangle—SSS and SAS. If you are given three sides (SSS), or two sides and their included angle (SAS), none of the ratios in the Law of Sines would be complete. In such cases, you can use the Law of Cosines.

Let's consider types of triangles with the three pieces of information shown below. We can't use the Law of Sines on these because we don't have an angle and a side opposite it. We need another method for SAS and SSS triangles. SAS AAA You may have a side, an angle, and then another side You may have all three angles. This case doesn't determine a triangle because similar triangles have the same angles and shape but "blown up" or "shrunk down" SSS You may have all three sides

Triangle Side Length Restriction In any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. Copyright © 2009 Pearson Addison-Wesley 1.1-5

Proof of the Law of Cosines Prove that c2= a2 + b2 – 2ab cos C In triangle CBD, cos C = x / a Then, x= a cos C (Eq #1) Using Pythagorean Theorem h2 = a2 - x2 (Eq #2) In triangle BDA, c2 = h2 + (b – x)2 c2 = h2 + b2 – 2bx + x2 (Eq #3) Substitute with h2 Eq #2 c2 = a2 - x2 + b2 – 2bx + x2 Combine like Terms c2 = a2 + b2 – 2bx Finally, substitute x with Eq #1 c2 = a2 + b2 – 2ba cos C Therefore: c2 = a2 + b2 – 2ab cos C C A B c a h b x D b - x Prove: c2 = a2 + b2 – 2ab cos C

Introduction

Using the Law of Cosines to Solve a Triangle (SAS) Example Solve triangle ABC if A = 42.3°, b = 12.9 meters, and c = 15.4 meters.

Using the Law of Cosines to Solve a Triangle (SAS) Angle B must be the smaller of the two remaining angles since it is opposite the shorter of the two sides b and c. Therefore, it cannot be obtuse. We use the Law of Sines to find angle B. 10.47 m 𝟖𝟏.𝟕° 𝟓𝟔.𝟎° Caution If we had chosen to find angle C rather than angle B, we would not have known whether angle C equals 81.7° or its supplement, 98.3°.

Law of Cosines Knowing the cosine of an angle, you can determine whether the angle is acute or obtuse. That is, cos  > 0 for 0 <  < 90 cos  < 0 for 90 <  < 180. If the largest angle is acute, the remaining two angles are acute also. Acute Obtuse

Using the Law of Cosines to Solve a Triangle (SAS) Step 1: Use the Law of Cosines to find the side opposite the given angle. Step 2: Use the Law of Sines to find the angle opposite the shorter of the two given sides. This angle is always acute. Step 3: Find the third angle. Subtract the measure of the given angle and the angle found in step 2 from 180°.

Using the Law of Cosines to Solve a Triangle (SSS) Example Solve triangle ABC if a = 9.47 ft, b =15.9 ft, and c = 21.1 ft. Solution We solve for C, the largest angle, first. If cos C < 0, then C will be obtuse. C 𝑏=15.9 109.9° 𝑎=9.47 B A 𝑐=21.1

Using the Law of Cosines to Solve a Triangle (SSS) We will use the Law of Sines to find B. sin 𝐵 15.9 = sin 109.9 21.1 sin 𝐵= 15.9∙ sin 109.9 21.1 sin 𝐵=.7086 sin −1 ( .7086)=45.1° C 𝑏=15.9 109.9° 𝑎=9.47 𝐴=180°−109.9°−45.1°≈25.0° 45.1° 25.0° B A 𝑐=21.1

Using the Law of Cosines to Solve a Triangle (SSS) Use the Law of Cosines to find the angle opposite the longest side. Use the Law of Sines to find either of the two remaining acute angles. Find the third angle by subtracting the measures of the angles found in steps 1 and 2 from 180°.

Example 1 USING THE LAW OF COSINES IN AN APPLICATION (SAS) A surveyor wishes to find the distance between two inaccessible points A and B on opposite sides of a lake. While standing at point C, she finds that AC = 259 m, BC = 423 m, and angle ACB measures 132°. Find the distance AB. Copyright © 2009 Pearson Addison-Wesley 1.1-15

The distance between the two points is about 627 m. Example 1 USING THE LAW OF COSINES IN AN APPLICATION (SAS) Use the law of cosines because we know the lengths of two sides of the triangle and the measure of the included angle. 𝐴𝐵 2 = 𝐴𝐶 2 + 𝐵𝐶 2 −2 𝐴𝐶 𝐵𝐶 cos 𝐶 𝐴𝐵 2 = 𝟐𝟓𝟗 𝟐 + 𝟒𝟐𝟑 𝟐 −2 𝟐𝟓𝟗 𝟒𝟐𝟑 cos 𝟏𝟑𝟐° 𝐴𝐵≈627 𝑚 The distance between the two points is about 627 m. Copyright © 2009 Pearson Addison-Wesley 1.1-16

When to use the Law of Sines and the Law of Cosines Four possible cases can occur when solving an oblique triangle. Copyright © 2009 Pearson Addison-Wesley 1.1-17

Copyright © 2009 Pearson Addison-Wesley 1.1-18

Copyright © 2009 Pearson Addison-Wesley 1.1-19