Stitz-Zeager_College_Algebra_e-book

# Once again we use the sum formula for cosine to get

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Unformatted text preview: SA case. 11.2 The Law of Sines 767 the Law of Sines gives sin(α) = sin(γ ) so that sin(γ ) = c sin(α) = a = 1. Here, γ = 90◦ as required. a c a a Moving along, now suppose h < a < c. As before, the Law of Sines9 gives sin(γ ) = c sin(α) . Since a h < a, c sin(α) < a or c sin(α) < 1 which means there are two solutions to sin(γ ) = c sin(α) : an a a acute angle which we’ll call γ0 , and its supplement, 180◦ − γ0 . We need to argue that each of these angles ‘ﬁt’ into a triangle with α. Since (α, a) and (γ0 , c) are angle-side opposite pairs, the assumption c > a in this case gives us γ0 > α. Since γ0 is acute, we must have that α is acute as well. This means one triangle can contain both α and γ0 , giving us one of the triangles promised in the theorem. If we manipulate the inequality γ0 > α a bit, we have 180◦ − γ0 < 180◦ − α which gives (180◦ − γ0 ) + α < 180◦ . This proves a triangle can contain both of the angles α and (180◦ − γ0 ), giving us the sec...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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