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Pre-Calc Exam Notes 4

Pre-Calc Exam Notes 4 - AB(that is CD forms a right angle...

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4 Chapter 1 Right Triangle Trigonometry §1.1 A C B b a c Figure 1.1.4 In a right triangle, the side opposite the right angle is called the hy- potenuse , and the other two sides are called its legs . For example, in Figure 1.1.4 the right angle is C , the hypotenuse is the line segment AB , which has length c , and BC and AC are the legs, with lengths a and b , respectively. The hypotenuse is always the longest side of a right triangle (see Exercise 11). By knowing the lengths of two sides of a right triangle, the length of the third side can be determined by using the Pythagorean Theorem : Theorem 1.1. Pythagorean Theorem: The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of its legs. Thus, if a right triangle has a hypotenuse of length c and legs of lengths a and b , as in Figure 1.1.4, then the Pythagorean Theorem says: a 2 + b 2 = c 2 (1.1) Let us prove this. In the right triangle ABC in Figure 1.1.5(a) below, if we draw a line segment from the vertex C to the point D on the hypotenuse such that CD is perpendicular to
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Unformatted text preview: AB (that is, CD forms a right angle with AB ), then this divides △ ABC into two smaller triangles △ CBD and △ ACD , which are both similar to △ ABC . A C B b a c D d c − (a) △ ABC C D B d a (b) △ CBD A D C c − d b (c) △ ACD Figure 1.1.5 Similar triangles △ ABC , △ CBD , △ ACD Recall that triangles are similar if their corresponding angles are equal, and that similarity implies that corresponding sides are proportional. Thus, since △ ABC is similar to △ CBD , by proportionality of corresponding sides we see that AB is to CB (hypotenuses) as BC is to BD (vertical legs) ⇒ c a = a d ⇒ cd = a 2 . Since △ ABC is similar to △ ACD , comparing horizontal legs and hypotenuses gives b c − d = c b ⇒ b 2 = c 2 − cd = c 2 − a 2 ⇒ a 2 + b 2 = c 2 . QED Note: The symbols ⊥ and ∼ denote perpendicularity and similarity, respectively. For exam-ple, in the above proof we had CD ⊥ AB and △ ABC ∼△ CBD ∼△ ACD ....
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