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Sec. 5.5 Law of sines.

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Presentation on theme: "Sec. 5.5 Law of sines."— Presentation transcript:

1 Sec. 5.5 Law of sines

2 Deriving the Law of Sines
C In either triangle: b a h In the top triangle: A B c C In the bottom triangle: b But, h a so each of these last two expressions are equal!!! A c B

3 Deriving the Law of Sines
C b a Solve for h: h A B c Set equal: C b Which is equivalent to: h a A c B

4 Law of Sines In any triangle with angles A, B, and C
opposite sides a, b, and c, respectively, the following equation is true: The Law of Sines works most easily with these two triangle cases: AAS, ASA

5 Guided Practice Solve , given the following. C b 8 A B c

6 Guided Practice Solve , given the following. C b 8 A B c

7 The Ambiguous Case (SSA)
We wish to construct ABC given angle A, side AB, & side BC. 1. Suppose angle A is obtuse and that side AB is as shown below. To complete the triangle, side BC must determine a point on the dotted horizontal line (which extends infinitely to the left). Explain from the picture why a unique triangle ABC is determined if BC > AB, but no triangle is determined if BC < AB. A B

8 The Ambiguous Case (SSA)
We wish to construct ABC given angle A, side AB, & side BC. 2. Suppose angle A is acute and that side AB is as shown below. To complete the triangle, side BC must determine a point on the dotted horizontal line (which extends infinitely to the right). Explain from the picture why a unique triangle ABC is determined if BC = h, but no triangle is determined if BC < h. B h A

9 The Ambiguous Case (SSA)
We wish to construct ABC given angle A, side AB, & side BC. 3. Suppose angle A is acute and that side AB is as shown below. If AB > BC > h, then we can form a triangle as shown. Find a second point C on the dotted horizontal line that gives a side BC of the same length, but determines a different triangle. (This is the “ambiguous case.”) B h A C C

10 The Ambiguous Case (SSA)
We wish to construct ABC given angle A, side AB, & side BC. 4. Explain why sin(C) is the same in both triangles in the ambiguous case. (This is why the Law of Sines is also ambiguous in this case.) 5. Explain from the figure below why a unique triangle is determined if BC > AB. B h A C C

11 Guided Practice Because h < c < b, one triangle is formed.
State whether the given measurements determine zero, one, or two triangles. (a) B = 82 , b = 17, c = 15 Solve for h: A 15 17 h 82 B C Because h < c < b, one triangle is formed.

12 Practice Problems Because a < h, no triangle is formed.
State whether the given measurements determine zero, one, or two triangles. (b) A = 73 , a = 24, b = 28 Solve for h: C 24 28 h 73 A B Because a < h, no triangle is formed.

13 Practice Problems Because h < c < a, two triangles are formed.
State whether the given measurements determine zero, one, or two triangles. (c) C = 31 , a = 17, c = 10 Solve for h: B 17 10 h 31 C A Because h < c < a, two triangles are formed.

14 Guided Practice (a) B = 38 , b = 21, c = 25
Two triangles can be formed using the given measurements. Solve both triangles. A Here, C is acute: (a) B = 38 , b = 21, c = 25 25 A 21 38 25 B 21 C 21 38 B C C

15 Guided Practice (a) B = 38 , b = 21, c = 25
Two triangles can be formed using the given measurements. Solve both triangles. (a) B = 38 , b = 21, c = 25 A A What if C is obtuse? 25 21 25 21 38 21 B C 38 B C C

16 Whiteboard Practice Solve , given the following. C 4 b A B c

17 Whiteboard Practice Solve , given the following. C b a A B 12

18 Whiteboard Practice Solve , given the following. C 28 32 A B c

19 Whiteboard Practice Solve , given the following. C a B 46 61 A

20 Whiteboard Problems (b) B = 57 , a = 11, b = 10
Two triangles can be formed using the given measurements. Solve both triangles. C Here, A is acute: (b) B = 57 , a = 11, b = 10 11 C 10 57 11 B 10 A 10 57 B A A

21 Whiteboard Problems (b) B = 57 , a = 11, b = 10
Two triangles can be formed using the given measurements. Solve both triangles. (b) B = 57 , a = 11, b = 10 C C What if A is obtuse? 11 10 11 10 57 10 B A 57 B A A


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