Pre-Calc Exam Notes 41

Pre-Calc Exam Notes 41 - C to AB has height h = b sin A ....

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The Law of Sines Section 2.1 41 In Example 2.2 we saw what is known as the ambiguous case . That is, there may be more than one solution. It is also possible for there to be exactly one solution or no solution at all. Example 2.3 Case 2: Two sides and one opposite angle. Solve the triangle ABC given a = 5, A = 30 , and b = 12. Solution: By the Law of Sines, we have sin B b = sin A a sin B = b sin A a = 12 sin 30 5 sin B = 1.2 , which is impossible since | sin B |≤ 1 for any angle B . Thus, there is no solution . There is a way to determine how many solutions a triangle has in Case 2. For a triangle ABC , suppose that we know the sides a and b and the angle A . Draw the angle A and the side b , and imagine that the side a is attached at the vertex at C so that it can “swing” freely, as indicated by the dashed arc in Figure 2.1.3 below. h b a A C B (a) a < h : No solution h b a c A C B (b) a = h : One solution h b a a A C B B (c) h < a < b : Two solutions b b a c A C B (d) a b : One solution Figure 2.1.3 The ambiguous case when A is acute If A is acute, then the altitude from
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Unformatted text preview: C to AB has height h = b sin A . As we can see in Figure 2.1.3(a)-(c), there is no solution when a &lt; h (this was the case in Example 2.3); there is exactly one solution - namely, a right triangle - when a = h ; and there are two solutions when h &lt; a &lt; b (as was the case in Example 2.2). When a b there is only one solution, even though it appears from Figure 2.1.3(d) that there may be two solutions, since the dashed arc intersects the horizontal line at two points. However, the point of intersection to the left of A in Figure 2.1.3(d) can not be used to determine B , since that would make A an obtuse angle, and we assumed that A was acute. If A is not acute (i.e. A is obtuse or a right angle), then the situation is simpler: there is no solution if a b , and there is exactly one solution if a &gt; b (see Figure 2.1.4)....
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