Introduction to Engineering Technology EIGHT EDITION CHAPTER 6 Right-Triangle Trigonometry and Geometry for Technologists Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Right-Triangle Relationships • Carpenters still use 3-4-5 Triangle • Pythagorean Theorem c a b 2 ► ► 2 2 Three angles total 180º For a right triangle, angles A and B total 90º Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Figure 6.2 The right triangle, in standard position, has sides and angles labeled in a conventional format. Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Trigonometric Functions opposite side sin hyponenuse adjacent side cos hyponenuse opposite side tan adjacent side Figure 6.4 The trigonometric functions are defined for the right triangle. Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Figure 6.5 Similar triangles have similar angles and the sin, cos, and tan ratios are equal. For instance, tan A = a'|b' = a"|b" = a'"|b'" = 1 (for angle A of 45°, as shown in the figure). Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Table 6.1 Partial Table of the Values for Trig Functions* Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Right – triangle Applications Example6.6: A triangular metal brace must be manufactured to support two welded pieces perpendicular to each other. If the longest side (hypotenuse) is 0.6m and the support side is 0.5m, solve for the two angles and the length of the base. Solution: First, check to be sure your calculator is in the degree mode. Solve for angle A by using the sine function: 𝑜𝑝𝑝 0.5 Sine A = ℎ𝑦𝑝 = 0.6 = 0.833 sine−1 0.833 = angle A = 56.4˚ Using the complementary relationship, angle B = 90.0˚- 56.4˚ = 33.6˚ Using Pythagorean theorem, 𝑏2 = 𝑐 2 - 𝑎2 A brace is to be manufactured to support a perpendicular weld b = 𝑐 2 − 𝑎2 = 0.36 − 0.25 = 0.332m In trigonometry, there are many ways to check your answers, and in this case you would use 𝑜𝑝𝑝 tan A = 𝑎𝑑𝑗 = b= 𝑜𝑝𝑝 tan 𝐴 = 0.5 tan 56.4˚ = 0.5 𝑏 0.5 1.505 = 0.332m The check back assures us that previous calculations were accurate. Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Figure 6.7 A brace is to be manufactured to support a perpendicular weld. Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Example 6.7: A surveyor checks the grade (slope) of a mountain highway. If the average grade is 8˚, what is the length of highway for each horizontal mile of distance? What is the percent grade? 𝑅𝑖𝑠𝑒 % grade = 𝑅𝑢𝑛 x 100 Solution: Because the adjacent side and angle A are known, the cosine function is selected: 𝑎𝑑𝑗 cos A = ℎ𝑦𝑝 𝑎𝑑𝑗 1 1 c(hyp) = cos 𝐴 = cos 8˚ = 0.990 = 1.01 mi Therefore, for every mile of horizontal distance covered, 1.01 mi must be asphalted. Use the Pythagorean theorem to find the rise, or side a. 𝑎2= 𝑐 2- 𝑏2 a = 1.012 − 12 = 0.142 mi The road rises 0.142mi for every mile of run (horizontal distance). The percent grade equals 𝑅𝑖𝑠𝑒 % grade = 𝑅𝑢𝑛 x 100 = Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen 0.142 1 = 14.2 Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Vector Applications Figure 6.12 Cross section of mountain for proposed tunnel. Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Figure 6.14 Two parallel lines cut by a transversal. The resulting alternate angles, alternate interior angles, and alternate exterior angles are equal. Alternate angles = A and D, B and C, E and H, F and G Alternate interior angles = C and F, D and E Alternate exterior angles = A and H, B and G Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Geometry Applications • • • • Circles Quadrilaterals Triangle Other polygons Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Circles Figure 6.17 The circumference of a circle, when divided by its diameter, yields a unitless ratio of 3.14. The Greek letter π (pi) represents this ratio, which applies to all circles. Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Quadrilaterals Figure 6.18 A few of the most familiar quadrilaterals found in business and industry. Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Triangles Figure 6.21 the height. All of the general types of triangles shown above have an area equal to one- half the base times Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved Solid Geometry Figure 6.24 Drawing of a right circular cylinder (A), the circular cylinder unwrapped (B), and the concept drawing for calculating the volume of the cylinder (C). Asurface = Atop and bottom + Aside Asurface = 2r2+ L X W Because W is equal to the circumference of the top or bottom and L to the height Asurface = 2r2+ 2r x h = 2r (r+h) Vcylinder = Atop or bottom x height = r2h Introduction to Engineering Technology, Eighth Edition Robert J. Pond | Jeffrey L. Rankinen Copyright © 2015 by Pearson Education, Inc. All Rights Reserved
