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.
Moving along, now suppose h < a < c. As before, the Law of Sines9 gives sin(γ ) = c sin(α) . Since
h < a, c sin(α) < a or c sin(α) < 1 which means there are two solutions to sin(γ ) = c sin(α) : an
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...
View Full Document