Pre-Calc Exam Notes 54

Pre-Calc Exam Notes 54 - Area = K = 1 2 bc sin A (2.23) The...

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54 Chapter 2 General Triangles §2.4 2.4 The Area of a Triangle In elementary geometry you learned that the area of a triangle is one-half the base times the height. We will now use that, combined with some trigonometry, to derive more formulas for the area when given various parts of the triangle. Case 1: Two sides and the included angle. Suppose that we have a triangle ABC , in which A can be either acute, a right angle, or obtuse, as in Figure 2.4.1. Assume that A , b , and c are known. h b a A C B c (a) A acute b h a A C B c (b) A = 90 h b a A C B c (c) A obtuse Figure 2.4.1 Area of ABC In each case we draw an altitude of height h from the vertex at C to AB , so that the area (which we will denote by the letter K ) is given by K = 1 2 hc . But we see that h = b sin A in each of the triangles (since h = b and sin A = sin 90 = 1 in Figure 2.4.1(b), and h = b sin (180 A ) = b sin A in Figure 2.4.1(c)). We thus get the following formula:
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Unformatted text preview: Area = K = 1 2 bc sin A (2.23) The above formula for the area of ABC is in terms of the known parts A , b , and c . Similar arguments for the angles B and C give us: Area = K = 1 2 ac sin B Area = K = 1 2 ab sin C (2.24) (2.25) Notice that the height h does not appear explicitly in these formulas, although it is implicitly there. These formulas have the advantage of being in terms of parts of the triangle, without having to nd h separately. Example 2.13 b = 5 a c = 7 A = 33 B C Find the area of the triangle ABC given A = 33 , b = 5, and c = 7. Solution: Using formula (2.23), the area K is given by: K = 1 2 bc sin A = 1 2 (5)(7) sin 33 K = 9.53...
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