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Transcript

MTH 250 Graded Assignment 9, part two 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 cos 45 csc 45 sec 45 tan 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] [So go ahead and do that. AD [Fill in 6]] 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. (Be sure you know them, obviously!)