1 / 16

Section 6.1.3 Day 1 The Ambiguous Case of the Law of Sines

Section 6.1.3 Day 1 The Ambiguous Case of the Law of Sines. Lesson Objective: Students will: Solve triangles that could have more than one solution. Extend their understanding of the inverse sine and cosine functions to triangles. Recall.

sabine
Download Presentation

Section 6.1.3 Day 1 The Ambiguous Case of the Law of Sines

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. Section 6.1.3 Day 1 The Ambiguous Case of the Law of Sines

  2. Lesson Objective: Students will: • Solve triangles that could have more than one solution. • Extend their understanding of the inverse sine and cosine functions to triangles.

  3. Recall • The Law of Sines provides a way to solve triangles that are not right triangles

  4. The Law of Sines is: C a b B c A

  5. Recall To use the Law of Sines, at least one angle and the measure of the side opposite that angle must be known.

  6. Law of Sines If given: • two angles and • the side opposite one of the given angles, Then: • one triangle exists and • use Law of Sines to find missing parts.

  7. Only One Angle Given When one angle and two sides (with one side across from the angle) are given, the following may be true: • no triangle exists • one triangle exists • two triangles exist

  8. Case 1 Given the measure of angle A is less than 90: • If a = bsinA then one triangle exists. • If a > bsinA and a > b one triangle exists. • If a > bsinA and a < b then 2 triangles exist. • If a < bSinA then no triangle exists.

  9. Case 2 Given the measure of angle A > 90: • If a < b then no triangleexists. • If a > b then one triangle exists.

  10. C 50sin60° 33 50 60° A B Example 1 Solve ABC if A=60° b=50 and a=33 a<bsinAnotriangle

  11. Example 2 C Solve ABC if A=60° b=50 and a=45 45 50 60° A B 2 Triangles

  12. Example 2 (cont.) C Solve ABC if A=60° b=50 and a=45 bsina 50 45 60° A B a>bsinA;a<b2 triangles

  13. Example 3 Solve ABC if A=60° b=50 and a=65 C bsina 50 65 60° A B 1 Triangle

  14. Example 3 (cont.) Solve ABC if A=60° b=50 and a=65 C bsina 50 65 60° A B a>bsina & a>b1 triangle

  15. Lesson Close What is the purpose of the Law of Sines?

  16. Assignment Board Problems #1-7

More Related