Mathematics For Machine Technology
Mathematics For Machine Technology
8th Edition
ISBN: 9781337798310
Author: Peterson, John.
Publisher: Cengage Learning,
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Chapter 56, Problem 20A

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 E F = 84 ° , find
(1) ∠EFD
(2) H F
(3) ∠1
b. If E F = 79 ° , find
(1) ∠EFD
(2) H F
(3) ∠1
Chapter 56, Problem 20A, Solve the following exercises based on Principles 18 through 21, although an exercise may require

Expert Solution
Check Mark
To determine

(a)

The value of EFD,HF,and 1 in the given figure.

Answer to Problem 20A

The value of EFD,HF,and 1 are

  DC=76.00°

  EOD=31.00°

  AC=134.00°

Explanation of Solution

Given information:

  EF=84°

The given figure is

  Mathematics For Machine Technology, Chapter 56, Problem 20A , additional homework tip  1

Calculation:

The angle drawn at the point of tangent between the tangent and the chord is equal to half the value of intercepted arc.

The angle EFD, is the angle between the tangent DF and the chord EF, thus, the value of angle EFD will be equal to

  EFD= EF2EFD=84°2=42°EFD=42°

The line DFP is the straight line, the angle at the straight line is 180o. The value of angle FHP is to be calculated for calculating the value of arc HF.

  EFD+EFH+HFP=180°42°+87°+HFP=180°HFP=180°(42°+87°)HFP=58°

The angle drawn at the point of tangent between the tangent and the chord is equal to half the value of intercepted arc. So,

  HFP= HF2HF=2×HFPHF=2×58°HF=116°

To calculate the value of angle 1, let us first calculate the value of arc FEH.

  EH=360°(79°+116°)EH=165°FEH=FE+EHFEH=79°+165°FEH=244°

Similarly, for angle 1, the angle drawn at the point of tangent between the tangent and the chord is equal to half the value of intercepted arc.

  1= FEH21=244°21=122°

Conclusion:

Thus, the value of EFD,HFand 1 are

  EFD=42°

  HF=116°

  1=122°

Expert Solution
Check Mark
To determine

(b)

The value of EFD,HF,and 1 in the given figure.

Answer to Problem 20A

The value of EFD,HF,and 1 are

  DC=76.00°

  EOD=31.00°

  AC=134.00°

Explanation of Solution

Given information:

  EF=84°

The given figure is

  Mathematics For Machine Technology, Chapter 56, Problem 20A , additional homework tip  2

Calculation:

The angle drawn at the point of tangent between the tangent and the chord is equal to half the value of intercepted arc.

The angle EFD, is the angle between the tangent DF and the chord EF, thus, the value of angle EFD will be equal to

  EFD= EF2EFD=84°2=42°EFD=42°

The line DFP is the straight line, the angle at the straight line is 180o. The value of angle FHP is to be calculated for calculating the value of arc HF.

  EFD+EFH+HFP=180°42°+87°+HFP=180°HFP=180°(42°+87°)HFP=58°

The angle drawn at the point of tangent between the tangent and the chord is equal to half the value of intercepted arc. So,

  HFP= HF2HF=2×HFPHF=2×58°HF=116°

To calculate the value of angle 1, let us first calculate the value of arc FEH.

  EH=360°(79°+116°)EH=165°FEH=FE+EHFEH=79°+165°FEH=244°

Similarly, for angle 1, the angle drawn at the point of tangent between the tangent and the chord is equal to half the value of intercepted arc.

  1= FEH21=244°21=122°

Conclusion:

Thus, the value of EFD,HFand 1 are

  EFD=42°

  HF=116°

  1=122°

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