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Example 5

One side of a triangle of length 20, forms a 420 angle with a second side. The length of the third side is 14. Find the length of the second side.


Let the required triangle be ABC. Given that a = 20, b = 14, and Ð B = 420

\ The first solution of D ABC is obtained.

The second solution will be obtained as

Let A' , B and C' and a' , b' and c' be the elements of the second solution. Let this triangle be A'BC'

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3. 1 Solving Right Triangles
3. 2 Law of Cosines
3. 3 Law of Sines
3. 4 The Ambiguous Case of Law of Sines
3. 5 Areas of Triangles
Supplementary Problems

Chapter 4

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