Solve the following exercises based on Principles 18 through 21, although an exercise may require the application of two 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 ∠ 1 = 67°00' and ∠ 2 =93°00', find: (1) A B ⌢ (2) D E ⌢ b.If ∠ 1 = 75°00' and ∠ 2 =85°00', find: (1) A B ⌢ (2) D E ⌢

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Answer 1

Textbook Question

Chapter 56, Problem 26A

Solve the following exercises based on Principles 18 through 21, although an exercise may require the application of two 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 ∠ 1 = 67°00' and ∠ 2 =93°00', find:
(1) A B ⌢
(2) D E ⌢
b.If ∠ 1 = 75°00' and ∠ 2 =85°00', find:
(1) A B ⌢
(2) D E ⌢
Chapter 56, Problem 26A, Solve the following exercises based on Principles 18 through 21, although an exercise may require

Expert Solution

To determine

(a)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'.

Explanation of Solution

Given information:

The value of angle 1 is ∠1=67°00'

The value of angle 2 is ∠2=93°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 1

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=67°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+67°=180°∠GFH=180°−(70°+67°)∠GFH=43°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=43°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×43°00'

AB⌢=86°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+93°00'=180°∠PMG=87°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+87°00'=180°∠MPG=180°−(70°00'+87°00')∠MPG=23°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢23°00'=62°00'−DE⌢DE⌢=62°00'−23°00'

DE⌢=39°00'

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Expert Solution

To determine

(b)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Explanation of Solution

Given information:

The value of angle 1 is ∠1=75°00'

The value of angle 2 is ∠2=85°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 2

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=75°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+75°=180°∠GFH=180°−(70°+75°)∠GFH=35°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=35°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×35°00'

AB⌢=70°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+85°00'=180°∠PMG=95°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+95°00'=180°∠MPG=180°−(70°00'+95°00')∠MPG=15°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢15°00'=62°00'−DE⌢DE⌢=62°00'−15°00'

DE⌢=47°00'

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

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Answer 2

Textbook Question

Chapter 56, Problem 26A

Solve the following exercises based on Principles 18 through 21, although an exercise may require the application of two 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 ∠ 1 = 67°00' and ∠ 2 =93°00', find:
(1) A B ⌢
(2) D E ⌢
b.If ∠ 1 = 75°00' and ∠ 2 =85°00', find:
(1) A B ⌢
(2) D E ⌢
Chapter 56, Problem 26A, Solve the following exercises based on Principles 18 through 21, although an exercise may require

Expert Solution

To determine

(a)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'.

Explanation of Solution

Given information:

The value of angle 1 is ∠1=67°00'

The value of angle 2 is ∠2=93°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 1

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=67°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+67°=180°∠GFH=180°−(70°+67°)∠GFH=43°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=43°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×43°00'

AB⌢=86°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+93°00'=180°∠PMG=87°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+87°00'=180°∠MPG=180°−(70°00'+87°00')∠MPG=23°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢23°00'=62°00'−DE⌢DE⌢=62°00'−23°00'

DE⌢=39°00'

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Expert Solution

To determine

(b)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Explanation of Solution

Given information:

The value of angle 1 is ∠1=75°00'

The value of angle 2 is ∠2=85°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 2

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=75°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+75°=180°∠GFH=180°−(70°+75°)∠GFH=35°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=35°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×35°00'

AB⌢=70°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+85°00'=180°∠PMG=95°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+95°00'=180°∠MPG=180°−(70°00'+95°00')∠MPG=15°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢15°00'=62°00'−DE⌢DE⌢=62°00'−15°00'

DE⌢=47°00'

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

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Answer 3

Textbook Question

Chapter 56, Problem 26A

Solve the following exercises based on Principles 18 through 21, although an exercise may require the application of two 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 ∠ 1 = 67°00' and ∠ 2 =93°00', find:
(1) A B ⌢
(2) D E ⌢
b.If ∠ 1 = 75°00' and ∠ 2 =85°00', find:
(1) A B ⌢
(2) D E ⌢
Chapter 56, Problem 26A, Solve the following exercises based on Principles 18 through 21, although an exercise may require

Expert Solution

To determine

(a)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'.

Explanation of Solution

Given information:

The value of angle 1 is ∠1=67°00'

The value of angle 2 is ∠2=93°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 1

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=67°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+67°=180°∠GFH=180°−(70°+67°)∠GFH=43°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=43°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×43°00'

AB⌢=86°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+93°00'=180°∠PMG=87°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+87°00'=180°∠MPG=180°−(70°00'+87°00')∠MPG=23°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢23°00'=62°00'−DE⌢DE⌢=62°00'−23°00'

DE⌢=39°00'

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Expert Solution

To determine

(b)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Explanation of Solution

Given information:

The value of angle 1 is ∠1=75°00'

The value of angle 2 is ∠2=85°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 2

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=75°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+75°=180°∠GFH=180°−(70°+75°)∠GFH=35°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=35°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×35°00'

AB⌢=70°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+85°00'=180°∠PMG=95°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+95°00'=180°∠MPG=180°−(70°00'+95°00')∠MPG=15°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢15°00'=62°00'−DE⌢DE⌢=62°00'−15°00'

DE⌢=47°00'

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

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Answer 4

Textbook Question

Chapter 56, Problem 26A

Solve the following exercises based on Principles 18 through 21, although an exercise may require the application of two 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 ∠ 1 = 67°00' and ∠ 2 =93°00', find:
(1) A B ⌢
(2) D E ⌢
b.If ∠ 1 = 75°00' and ∠ 2 =85°00', find:
(1) A B ⌢
(2) D E ⌢
Chapter 56, Problem 26A, Solve the following exercises based on Principles 18 through 21, although an exercise may require

Expert Solution

To determine

(a)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'.

Explanation of Solution

Given information:

The value of angle 1 is ∠1=67°00'

The value of angle 2 is ∠2=93°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 1

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=67°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+67°=180°∠GFH=180°−(70°+67°)∠GFH=43°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=43°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×43°00'

AB⌢=86°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+93°00'=180°∠PMG=87°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+87°00'=180°∠MPG=180°−(70°00'+87°00')∠MPG=23°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢23°00'=62°00'−DE⌢DE⌢=62°00'−23°00'

DE⌢=39°00'

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Expert Solution

To determine

(b)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Explanation of Solution

Given information:

The value of angle 1 is ∠1=75°00'

The value of angle 2 is ∠2=85°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 2

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=75°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+75°=180°∠GFH=180°−(70°+75°)∠GFH=35°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=35°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×35°00'

AB⌢=70°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+85°00'=180°∠PMG=95°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+95°00'=180°∠MPG=180°−(70°00'+95°00')∠MPG=15°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢15°00'=62°00'−DE⌢DE⌢=62°00'−15°00'

DE⌢=47°00'

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'.

Explanation of Solution

Given information:

The value of angle 1 is ∠1=67°00'

The value of angle 2 is ∠2=93°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 1

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=67°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+67°=180°∠GFH=180°−(70°+67°)∠GFH=43°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=43°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×43°00'

AB⌢=86°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+93°00'=180°∠PMG=87°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+87°00'=180°∠MPG=180°−(70°00'+87°00')∠MPG=23°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢23°00'=62°00'−DE⌢DE⌢=62°00'−23°00'

DE⌢=39°00'

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Expert Solution

To determine

(b)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Explanation of Solution

Given information:

The value of angle 1 is ∠1=75°00'

The value of angle 2 is ∠2=85°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 2

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=75°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+75°=180°∠GFH=180°−(70°+75°)∠GFH=35°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=35°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×35°00'

AB⌢=70°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+85°00'=180°∠PMG=95°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+95°00'=180°∠MPG=180°−(70°00'+95°00')∠MPG=15°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢15°00'=62°00'−DE⌢DE⌢=62°00'−15°00'

DE⌢=47°00'

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Answer 8

Expert Solution

To determine

(a)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'.

Explanation of Solution

Given information:

The value of angle 1 is ∠1=67°00'

The value of angle 2 is ∠2=93°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 1

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=67°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+67°=180°∠GFH=180°−(70°+67°)∠GFH=43°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=43°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×43°00'

AB⌢=86°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+93°00'=180°∠PMG=87°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+87°00'=180°∠MPG=180°−(70°00'+87°00')∠MPG=23°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢23°00'=62°00'−DE⌢DE⌢=62°00'−23°00'

DE⌢=39°00'

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer 13

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'.

Answer 14

Explanation of Solution

Given information:

The value of angle 1 is ∠1=67°00'

The value of angle 2 is ∠2=93°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 1

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=67°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+67°=180°∠GFH=180°−(70°+67°)∠GFH=43°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=43°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×43°00'

AB⌢=86°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+93°00'=180°∠PMG=87°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+87°00'=180°∠MPG=180°−(70°00'+87°00')∠MPG=23°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢23°00'=62°00'−DE⌢DE⌢=62°00'−23°00'

DE⌢=39°00'

The values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=86°00' and DE⌢=39°00'

Answer 15

Expert Solution

To determine

(b)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Explanation of Solution

Given information:

The value of angle 1 is ∠1=75°00'

The value of angle 2 is ∠2=85°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 2

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=75°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+75°=180°∠GFH=180°−(70°+75°)∠GFH=35°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=35°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×35°00'

AB⌢=70°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+85°00'=180°∠PMG=95°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+95°00'=180°∠MPG=180°−(70°00'+95°00')∠MPG=15°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢15°00'=62°00'−DE⌢DE⌢=62°00'−15°00'

DE⌢=47°00'

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

The value of arc AB⌢ and DE⌢ in the given figure.

Answer 20

Answer to Problem 26A

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Answer 21

Explanation of Solution

Given information:

The value of angle 1 is ∠1=75°00'

The value of angle 2 is ∠2=85°00'

The given figure is

Mathematics For Machine Technology, Chapter 56, Problem 26A , additional homework tip 2

Calculation:

The angles GHF and angle 1 are opposite angles. Thus,

∠GHF=∠BHC∠GHF=∠1∠GHF=75°00'

Similarly, the angles EGM and angle FGH are opposite angles.

∠EGM=∠FGH70°00'=∠FGH∠FGH=70°00'

In triangle FGH, the angle GFH comes out to be

∠GFH+∠FGH+∠GHF=180°∠GFH+70°+75°=180°∠GFH=180°−(70°+75°)∠GFH=35°

Now, an inscribed angle is one half the value of intercepted arc by the angle itself.

∠GFH=∠AFB=35°00'∠AFB=12( AB⌢)AB⌢=2×∠AFBAB⌢=2×35°00'

AB⌢=70°00'

The line PMB is a straight line so,

∠PMG+∠GMB=180°∠PMG+85°00'=180°∠PMG=95°00'

Now, in triangle PMG, the sum of all angles should be equal to 180^o.

∠MPG+∠PGM+∠PMG=180°∠MPG+70°00'+95°00'=180°∠MPG=180°−(70°00'+95°00')∠MPG=15°00'

Now, the value of arc DE can be found from the following formula,

∠P=BC⌢−DE⌢15°00'=62°00'−DE⌢DE⌢=62°00'−15°00'

DE⌢=47°00'

The values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Conclusion:

Thus, the values of arcs AB and DE are AB⌢=70°00' and DE⌢=47°00'

Answer 22

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Answer to Problem 26A

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