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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.
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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 • The Law of Sines provides a way to solve triangles that are not right triangles
The Law of Sines is: C a b B c A
Recall To use the Law of Sines, at least one angle and the measure of the side opposite that angle must be known.
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.
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
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.
Case 2 Given the measure of angle A > 90: • If a < b then no triangleexists. • If a > b then one triangle exists.
C 50sin60° 33 50 60° A B Example 1 Solve ABC if A=60° b=50 and a=33 a<bsinAnotriangle
Example 2 C Solve ABC if A=60° b=50 and a=45 45 50 60° A B 2 Triangles
Example 2 (cont.) C Solve ABC if A=60° b=50 and a=45 bsina 50 45 60° A B a>bsinA;a<b2 triangles
Example 3 Solve ABC if A=60° b=50 and a=65 C bsina 50 65 60° A B 1 Triangle
Example 3 (cont.) Solve ABC if A=60° b=50 and a=65 C bsina 50 65 60° A B a>bsina & a>b1 triangle
Lesson Close What is the purpose of the Law of Sines?
Assignment Board Problems #1-7