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Exact values for trigonometric ratios The 30° − 60° − 90° and 45° − 45° − 90° triangles are familiar tools used to work out the exact values of the trig ratios for those angles (the idea being that, instead of memorizing all the ratios for 30°, 60°, and 45° , you memorize the proportions on the triangles, do a quick sketch, and then work out the ratios from there using the definitions). The goal here is to derive/prove that those lengths associated with the angles are correct. 45°− 45°− 90° Start by constructing a right triangle on a 45° angle, with side length 1. You can take the construction as a given; we won't justify all the steps that go into proving it's constructable (this is essentially segment construction to measure out side length, existence of a perpendicular, and three noncollinear points determining a triangle). By the Euclidean angle sum theorem, the value of (B must be [Fill in 1]. Since [Fill in 2 – what angles are congruent?] this is an [Fill in 3] triangle with base AB . [Fill in 4 – justification]. Since the triangle is [Fill in 3 again], we know that [Fill in 5 – sides?] by [Fill in 6 – justification], and therefore BC = 1 . Finally, can use the [Fill in 7 – theorem] to calculate the length AC = 2 . This (plus the definitions of the trig ratios) allows us to calculate the following values: sin 45° = csc 45° = cos 45° = tan 45° = sec 45° = cot 45° = 30°− 60°− 90° Start by constructing an equilateral triangle of side length 2. In Euclidean geometry, equilateral triangles are equiangular, so the values of all the angles are 60° . Construct an altitude (by definition, this is perpendicular to the base). [1] Write a short proof that + BAD ≅+CAD : [2] Use this to prove that AD bisects both the segment BC and the angle (BAC . Therefore, BD = [Fill in 3] and m(BAD = [Fill in 4]. The remaining length AD can be computed using the [Fill in 5] [6 : So go ahead and do that.] Note that the values of the trigonometric ratios for 30° and 60° were already computed in the notes, using the proportions from the 30° − 60° − 90° triangle, so you don't have to do that bit again.