1 / 11

The Law of Cosines

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.

samgonzales
Download Presentation

The Law of Cosines

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The Law of Cosines

  2. 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

  3. LAW OF COSINES Law of Cosines Use these to findmissing sides and angles C B A C a b B A c

  4. B c a h C A b 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, Using Pythagorean Theorem 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- x2+ b2 – 2bx + x2 c2= a2 + b2 – 2bx Finally, substitute x with Eq #1 c2= a2 + b2 – 2ba cos C Therefore: c2 = a2 + b2 – 2ab cos C D x b - x Prove: c2 = a2 + b2 – 2ab cos C

  5. Example 1. Solve a triangle where b = 1, c = 3 and A = 80° B 3 Draw a picture. a 80° A C This is SAS 1 A minus 2 times the productof those other sides times the cosine of the angle between those sides One side squared sum of each of the other sides squared Do we know an angle and side opposite it? No, so we must use Law of Cosines. Hint: we will be solving for the side opposite the angle we know.

  6. Example 1.Solve a triangle where b = 1, c = 3 and A = 80° B a 3 C 80° A 1 A minus 2 times the productof those other sides times the cosine of the angle between those sides sum of each of the other sides squared One side squared a = 2.99

  7. We'll label side a with the value we found. B We now have all of the sides but how can we find an angle?  3 18.85° 2.99  80° 81.15° Hint: We have an angle and a side opposite it. A C 1 C Angle  is easy to find since the sum of the angles is a triangle is 180° C 180 - 80- 81.15 = 18.85 B = 18.85° C = 81.15°

  8. Example #2 Solve a triangle where a = 5, b = 8 and c = 9 This isSSS Draw a picture. B Do we know an angle and side opposite it? No, so we must use Law of Cosines. 9 5 84.3 C A 8 C Let's use largest side to find largest angle first.

  9. Example #2 B 9 5 C = 84.3° 84.3  A C 8 C minus 2 times the productof those other sides times the cosine of the angle between those sides sum of each of the other sides squared One side squared C C C C

  10. Example #2 B  9 62.2 Do we know an angle and side opposite it? 5   84.3 33.5 A C 8 Yes, so use Law of Sines. B A A = 33.5° B B B = 62.2°

  11. Solve: • 1.) b = 40, c = 45, A = 51° a = 36.87, B = 57.47°, C = 71.53° • 2. ) B= 42°, a= 120, c=160 b = 107.07, A = 48.58°, C = 89.42° • 3. ) a =20, b= 12, c = 28 B = 21.79°, A = 89.79°, C = 120° • 4.) a= 15, b= 18, c = 17 B = 68.12°, A = 50.66°, C = 61.22°

More Related