EBK MATHEMATICS FOR MACHINE TECHNOLOGY
EBK MATHEMATICS FOR MACHINE TECHNOLOGY
7th Edition
ISBN: 9780100548169
Author: SMITH
Publisher: YUZU
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Chapter 54, Problem 22A

Solve the following exercises based on Principles 18 through 21, although an exercise may require the application oftwo or more of any of the principles. Where necessary, round linear answers in inches to 3 decimal places and millimeters to 2 decimal places. Round angular answers in decimal degrees to 2 decimal places and degrees and minutes to the nearest minute.

a. If A B = 72 ° 18 ' and C D = 50 ° 24 ' find:
(1) ∠1
(2) ∠2
(3) ∠3
b. If C D = 43 ° 15 ' and A D = 106 ° 05 ' , find:
(1) ∠1
(2) ∠2
(3) ∠3
Chapter 54, Problem 22A, Solve the following exercises based on Principles 18 through 21, although an exercise may require

Expert Solution
Check Mark
To determine

(a)

The values of angles 1,2and 3

Answer to Problem 22A

The values of angles are

  1=50°57'

  2=53°30'

  3=75°32'

Explanation of Solution

Given information:

  AB=72°20'CD=50°18'

Given figure is

EBK MATHEMATICS FOR MACHINE TECHNOLOGY, Chapter 54, Problem 22A , additional homework tip  1

Calculation:

The angle which is formed at a point on outside a circle by two tangents, two secants or a secant and a tangent is equal to one half the difference of arcs intercepted by the tangents or secants.

Let us first calculate the value of arc AD,

The total angle subtended by the entire circular arc on the center is 360o.

  AD+DC+BC+AB=360°AD+50°18'+108°19'+72°20'=360°AD=360°(50°18'+108°19'+72°20')

  AD=129°3'

Now,

  1=12( ABCD AD)Putting the given values1=12(( 50°18'+108°19'+72°20') 129°3')1=12(( 230°57')129°3')1=12×(101°54')

  1=50°57'

Now, for calculating angle 2,

Similarly,

  2=12( ADC AB)Putting the given values2=12(( AD + DC ) AB)2=12(( 129°3')+50°18')72°20'2=12×107°1'

  2=53°30'

Now, for calculating angle 3,

Similarly,

  3=12( DAB DC)Putting the given values3=12(( AD + AB ) DC)3=12(( 129°3')+72°20')50°18'

  3=75°32'

Conclusion:

Thus, the values of angles are

  1=50°57'

  2=53°30'

  3=75°32'

Expert Solution
Check Mark
To determine

(b)

The values of angles 1,2and 3

Answer to Problem 22A

The values of angles are

  1=73°55'

  2=23°30'

  3=82°36'

Explanation of Solution

Given information:

  AD=106°05'CD=43°15'

Given figure is

EBK MATHEMATICS FOR MACHINE TECHNOLOGY, Chapter 54, Problem 22A , additional homework tip  2

Calculation:

The angle which is formed at a point on outside a circle by two tangents, two secants or a secant and a tangent is equal to one half the difference of arcs intercepted by the tangents or secants.

Let us first calculate the value of arc AB,

The total angle subtended by the entire circular arc on the center is 360o.

  AD+DC+BC+AB=360°106°05'+43°15'+108°19'+AB=360°AB=360°(106°05'+43°15'+108°19')AB=360°257°39'

  AB=102°21'

Now,

  1=12( ABCD AD)Putting the given values1=12(( 102°21'+108°19'+43°15')106°05')1=12(( 253°55')106°05')1=12×(147°50')

  1=73°55'

Now, for calculating angle 2,

Similarly,

  2=12( ADC AB)Putting the given values2=12(( AD + DC ) AB)2=12(( 106°05')+43°15')102°21'2=12×46°59'

  2=23°30'

Now, for calculating angle 3,

Similarly,

  3=12( DAB DC)Putting the given values3=12(( AD + AB ) DC)3=12(( 106°05')+102°21')43°15'3=12×165°11'

  3=82°36'

Conclusion:

Thus, the values of angles are

  1=73°55'

  2=23°30'

  3=82°36'

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Chapter 54 Solutions

EBK MATHEMATICS FOR MACHINE TECHNOLOGY

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