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Pre-Calc Exam Notes 3

Pre-Calc Exam Notes 3 - ACB is a right angle(see Figure...

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Angles Section 1.1 3 Example 1.1 For each triangle below, determine the unknown angle(s): A B C 35 20 D E F 53 X Y Z α α 3 α Note: We will sometimes refer to the angles of a triangle by their vertex points. For example, in the ±rst triangle above we will simply refer to the angle BAC as angle A . Solution: For triangle ABC , A = 35 and C = 20 , and we know that A + B + C = 180 , so 35 + B + 20 = 180 B = 180 35 20 B = 125 . For the right triangle DEF , E = 53 and F = 90 , and we know that the two acute angles D and E are complementary, so D + E = 90 D = 90 53 D = 37 . For triangle XY Z , the angles are in terms of an unknown number α , but we do know that X + Y + Z = 180 , which we can use to solve for α and then use that to solve for X , Y , and Z : α + 3 α + α = 180 5 α = 180 α = 36 X = 36 , Y = 3 × 36 = 108 , Z = 36 Example 1.2 Thales’ Theorem states that if A , B , and C are (distinct) points on a circle such that the line segment AB is a diameter of the circle, then the angle
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Unformatted text preview: ACB is a right angle (see Figure 1.1.3(a)). In other words, the triangle △ ABC is a right triangle. A B C O (a) A B C O α α β β (b) Figure 1.1.3 Thales’ Theorem: ∠ ACB = 90 ◦ To prove this, let O be the center of the circle and draw the line segment OC , as in Figure 1.1.3(b). Let α = ∠ BAC and β = ∠ ABC . Since AB is a diameter of the circle, OA and OC have the same length (namely, the circle’s radius). This means that △ OAC is an isosceles triangle, and so ∠ OCA = ∠ OAC = α . Likewise, △ OBC is an isosceles triangle and ∠ OCB = ∠ OBC = β . So we see that ∠ ACB = α + β . And since the angles of △ ABC must add up to 180 ◦ , we see that 180 ◦ = α + ( α + β ) + β = 2( α + β ), so α + β = 90 ◦ . Thus, ∠ ACB = 90 ◦ . QED...
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