Explanation Given: A right triangle ABC with ∠ A = θ and | A C | = b .CD is drawn perpendicular to AB, DE is perpendicular to BC, EF perpendicular to AB and so on as shown in figure below: Definition Used: For a right angled triangle with an angle θ , sin θ = P e r p e n d i c u l a r H y p o t e n u s e Formula Used: For the infinite geometric series a , a r , a r 2 , ... , the sum of its terms is given by S = a 1 − r where 0 < r < 1 Now it is given that AC, DE, FF and so on are all perpendicular to the base BC of the triangle ABC. So, all these are parallel to each other. It implies that DC, FE, HG and so on are transversals for the parallel lines. Now, it is known that the alternate angles of two parallel lines are equal. Since AC is parallel to DE and DC is a transversal, so this implies that, ∠ A C D = ∠ E D C Similarly, ∠ D E F = ∠ G F E and so on Now, by the angle sum property of triangle, sum of all its angles are equal to 180 ° So, in triangle ADC ∠ A + ∠ A C D + ∠ A D C = 180 ° θ + ∠ A C D + 90 ° = 180 ° ∠ A C D = 90 ° − θ Applying angle sum property on triangle DEC and using ∠ A C D = ∠ E D C , ∠ E D C + ∠ D C E + ∠ D E C = 180 ° ∠ A C D + ∠ D C E + 90 ° = 180 ° 90 ° − θ + ∠ D C E + 90 ° = 180 ° ∠ D C E = θ Thus, it can be said that, θ = ∠ A = ∠ D C E = ∠ F D E = ∠ F E G = ⋯ Now, consider the right triangle ADC and use the definition of sin θ , sin θ = | D C | | A C | sin θ

Bundle: Biocalculus: Calculus, Probability, And Statistics For The Life Sciences + Webassign Printed Access Card For Stewart/day's Biocalculus: Calculus For Life Sciences, 1st Edition, Multi-term

1st Edition

ISBN: 9781305779037

Author: James Stewart, Troy Day

Publisher: Cengage Learning

See similar textbooks

Videos

Question

Chapter 2.1, Problem 60E

To determine

To find:total length of all the perpendiculars |CD|+|DE|+|EF|+|FG|+⋯ in terms of b and θ .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 2.1, Problem 59E

Chapter 2.1, Problem 1P

Students have asked these similar questions

The bracket BCD is hinged at C and attached to a control cable at B. Let F₁ = 275 N and F2 = 275 N. F1 B a=0.18 m C A 0.4 m -0.4 m- 0.24 m Determine the reaction at C. The reaction at C N Z F2 D

The correct answer is C,i know that we need to use stokes theorem and parametrize the equations then write the equation F with respect to the curve but i cant seem to find a way to do it, the integral should be from 0 to 2pi but i might be wrongcould you show me the steps to get to 18pi

A 10-ft boom is acted upon by the 810-lb force as shown in the figure. D 6 ft 6 ft E B 7 ft C 6 ft 4 ft W Determine the tension in each cable and the reaction at the ball-and-socket joint at A. The tension in cable BD is lb. The tension in cable BE is lb. The reaction at A is ( lb) i + Ib) j. (Include a minus sign if necessary.)

Chapter 2 Solutions

Bundle: Biocalculus: Calculus, Probability, And Statistics For The Life Sciences + Webassign Printed Access Card For Stewart/day's Biocalculus: Calculus For Life Sciences, 1st Edition, Multi-term

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E

Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - Prob. 41E Ch. 2.1 - Prob. 42E Ch. 2.1 - Prob. 43E Ch. 2.1 - Prob. 44E Ch. 2.1 - Prob. 45E Ch. 2.1 - Prob. 46E Ch. 2.1 - Prob. 47E Ch. 2.1 - Prob. 48E Ch. 2.1 - Prob. 49E Ch. 2.1 - Prob. 50E Ch. 2.1 - Prob. 51E Ch. 2.1 - Prob. 52E Ch. 2.1 - Prob. 53E Ch. 2.1 - Prob. 54E Ch. 2.1 - Prob. 55E Ch. 2.1 - Prob. 56E Ch. 2.1 - Prob. 57E Ch. 2.1 - Prob. 58E Ch. 2.1 - Prob. 59E Ch. 2.1 - Prob. 60E Ch. 2.1 - Prob. 1P Ch. 2.1 - Prob. 2P Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - Prob. 7P Ch. 2.1 - Prob. 8P Ch. 2.1 - Prob. 9P Ch. 2.2 - Prob. 1E Ch. 2.2 - Prob. 2E Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - Prob. 9E Ch. 2.2 - Prob. 10E Ch. 2.2 - Prob. 11E Ch. 2.2 - Prob. 12E Ch. 2.2 - Prob. 13E Ch. 2.2 - Prob. 14E Ch. 2.2 - Prob. 15E Ch. 2.2 - Prob. 16E Ch. 2.2 - Prob. 17E Ch. 2.2 - Prob. 18E Ch. 2.2 - Prob. 19E Ch. 2.2 - Prob. 20E Ch. 2.2 - Prob. 21E Ch. 2.2 - Prob. 22E Ch. 2.2 - Prob. 23E Ch. 2.2 - Prob. 24E Ch. 2.2 - Prob. 25E Ch. 2.2 - Prob. 26E Ch. 2.2 - Prob. 27E Ch. 2.2 - Prob. 28E Ch. 2.2 - Prob. 29E Ch. 2.2 - Prob. 30E Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.2 - Prob. 33E Ch. 2.2 - Prob. 34E Ch. 2.2 - Prob. 35E Ch. 2.2 - Prob. 36E Ch. 2.2 - Prob. 37E Ch. 2.3 - Prob. 1E Ch. 2.3 - Prob. 2E Ch. 2.3 - Prob. 3E Ch. 2.3 - Prob. 4E Ch. 2.3 - Prob. 5E Ch. 2.3 - Prob. 6E Ch. 2.3 - Prob. 7E Ch. 2.3 - Prob. 8E Ch. 2.3 - Prob. 9E Ch. 2.3 - Prob. 10E Ch. 2.3 - Prob. 11E Ch. 2.3 - Prob. 12E Ch. 2.3 - Prob. 13E Ch. 2.3 - Prob. 14E Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 16E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - Prob. 21E Ch. 2.3 - Prob. 22E Ch. 2.3 - Prob. 23E Ch. 2.3 - Prob. 24E Ch. 2.3 - Prob. 25E Ch. 2.3 - Prob. 26E Ch. 2.3 - Prob. 27E Ch. 2.3 - Prob. 28E Ch. 2.3 - Prob. 29E Ch. 2.3 - Prob. 30E Ch. 2.3 - Prob. 31E Ch. 2.3 - Prob. 32E Ch. 2.3 - Prob. 33E Ch. 2.3 - Prob. 34E Ch. 2.3 - Prob. 35E Ch. 2.3 - Prob. 36E Ch. 2.3 - Prob. 37E Ch. 2.3 - Prob. 38E Ch. 2.3 - Prob. 39E Ch. 2.3 - Prob. 40E Ch. 2.3 - Prob. 41E Ch. 2.3 - Prob. 42E Ch. 2.3 - Prob. 43E Ch. 2.3 - Prob. 44E Ch. 2.4 - Prob. 1E Ch. 2.4 - Prob. 2E Ch. 2.4 - Prob. 3E Ch. 2.4 - Prob. 4E Ch. 2.4 - Prob. 5E Ch. 2.4 - Prob. 6E Ch. 2.4 - Prob. 7E Ch. 2.4 - Prob. 8E Ch. 2.4 - Prob. 9E Ch. 2.4 - Prob. 10E Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - Prob. 14E Ch. 2.4 - Prob. 15E Ch. 2.4 - Prob. 16E Ch. 2.4 - Prob. 17E Ch. 2.4 - Prob. 18E Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - Prob. 21E Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - Prob. 24E Ch. 2.4 - Prob. 25E Ch. 2.4 - Prob. 26E Ch. 2.4 - Prob. 27E Ch. 2.4 - Prob. 28E Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Prob. 32E Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.4 - Prob. 37E Ch. 2.4 - Prob. 38E Ch. 2.4 - Prob. 39E Ch. 2.4 - Prob. 40E Ch. 2.4 - Prob. 41E Ch. 2.4 - Prob. 42E Ch. 2.4 - Prob. 43E Ch. 2.4 - Prob. 44E Ch. 2.4 - Prob. 45E Ch. 2.4 - Prob. 46E Ch. 2.4 - Prob. 47E Ch. 2.4 - Prob. 48E Ch. 2.4 - Prob. 49E Ch. 2.4 - Prob. 50E Ch. 2.4 - Prob. 51E Ch. 2.4 - Prob. 52E Ch. 2.5 - Prob. 1E Ch. 2.5 - Prob. 2E Ch. 2.5 - Prob. 3E Ch. 2.5 - Prob. 4E Ch. 2.5 - Prob. 5E Ch. 2.5 - Prob. 6E Ch. 2.5 - Prob. 7E Ch. 2.5 - Prob. 8E Ch. 2.5 - Prob. 9E Ch. 2.5 - Prob. 10E Ch. 2.5 - Prob. 11E Ch. 2.5 - Prob. 12E Ch. 2.5 - Prob. 13E Ch. 2.5 - Prob. 14E Ch. 2.5 - Prob. 15E Ch. 2.5 - Prob. 16E Ch. 2.5 - Prob. 17E Ch. 2.5 - Prob. 18E Ch. 2.5 - Prob. 19E Ch. 2.5 - Prob. 20E Ch. 2.5 - Prob. 21E Ch. 2.5 - Prob. 22E Ch. 2.5 - Prob. 23E Ch. 2.5 - Prob. 24E Ch. 2.5 - Prob. 25E Ch. 2.5 - Prob. 26E Ch. 2.5 - Prob. 27E Ch. 2.5 - Prob. 28E Ch. 2.5 - Prob. 29E Ch. 2.5 - Prob. 30E Ch. 2.5 - Prob. 31E Ch. 2.5 - Prob. 32E Ch. 2.5 - Prob. 33E Ch. 2.5 - Prob. 34E Ch. 2.5 - Prob. 35E Ch. 2.5 - Prob. 36E Ch. 2.5 - Prob. 37E Ch. 2.5 - Prob. 38E Ch. 2.5 - Prob. 39E Ch. 2.5 - Prob. 40E Ch. 2.5 - Prob. 41E Ch. 2.5 - Prob. 42E Ch. 2.5 - Prob. 43E Ch. 2.5 - Prob. 44E Ch. 2.5 - Prob. 45E Ch. 2.5 - Prob. 46E Ch. 2.5 - Prob. 47E Ch. 2.5 - Prob. 48E Ch. 2.5 - Prob. 49E Ch. 2.5 - Prob. 50E Ch. 2.5 - Prob. 51E Ch. 2 - Prob. 1CC Ch. 2 - Prob. 2CC Ch. 2 - Prob. 3CC Ch. 2 - Prob. 4CC Ch. 2 - Prob. 5CC Ch. 2 - Prob. 6CC Ch. 2 - Prob. 7CC Ch. 2 - Prob. 8CC Ch. 2 - Prob. 9CC Ch. 2 - Prob. 10CC Ch. 2 - Prob. 1TFQ Ch. 2 - Prob. 2TFQ Ch. 2 - Prob. 3TFQ Ch. 2 - Prob. 4TFQ Ch. 2 - Prob. 5TFQ Ch. 2 - Prob. 6TFQ Ch. 2 - Prob. 7TFQ Ch. 2 - Prob. 8TFQ Ch. 2 - Prob. 9TFQ Ch. 2 - Prob. 10TFQ Ch. 2 - Prob. 11TFQ Ch. 2 - Prob. 12TFQ Ch. 2 - Prob. 13TFQ Ch. 2 - Prob. 14TFQ Ch. 2 - Prob. 15TFQ Ch. 2 - Prob. 16TFQ Ch. 2 - Prob. 1E Ch. 2 - Prob. 2E Ch. 2 - Prob. 3E Ch. 2 - Prob. 4E Ch. 2 - Prob. 5E Ch. 2 - Prob. 6E Ch. 2 - Prob. 7E Ch. 2 - Prob. 8E Ch. 2 - Prob. 9E Ch. 2 - Prob. 10E Ch. 2 - Prob. 11E Ch. 2 - Prob. 12E Ch. 2 - Prob. 13E Ch. 2 - Prob. 14E Ch. 2 - Prob. 15E Ch. 2 - Prob. 16E Ch. 2 - Prob. 17E Ch. 2 - Prob. 18E Ch. 2 - Prob. 19E Ch. 2 - Prob. 20E Ch. 2 - Prob. 21E Ch. 2 - Prob. 22E Ch. 2 - Prob. 23E Ch. 2 - Prob. 24E Ch. 2 - Prob. 25E Ch. 2 - Prob. 26E Ch. 2 - Prob. 27E Ch. 2 - Prob. 28E Ch. 2 - Prob. 29E Ch. 2 - Prob. 30E Ch. 2 - Prob. 31E Ch. 2 - Prob. 32E Ch. 2 - Prob. 33E Ch. 2 - Prob. 34E Ch. 2 - Prob. 1CS Ch. 2 - Prob. 2CS Ch. 2 - Prob. 3CS Ch. 2 - Prob. 4CS Ch. 2 - Prob. 5CS Ch. 2 - Prob. 6CS Ch. 2 - Prob. 7CS Ch. 2 - Prob. 8CS Ch. 2 - Prob. 9CS

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

total length of all the perpendiculars | C D | + | D E | + | E F | + | F G | + ⋯ in terms of b and θ .

Solution Summary: The author explains that AC, DE, FF and so on are all perpendicular to the base BC of the triangle ABC. Since AC is parallel to DE and DC is a transversal, the alternate angles of two

Videos

To find:total length of all the perpendiculars |CD|+|DE|+|EF|+|FG|+⋯ in terms of b and θ .

Want to see the full answer?

Chapter 2 Solutions