Use the angle values given to find ∠2. a. If ∠1 = 37°, find ∠2. b. If ∠1 = 29°, find ∠1.

Question

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Answer 1

Textbook Question

Chapter 54, Problem 30A

Use the angle values given to find ∠2.
a. If ∠1 = 37°, find ∠2.
b. If ∠1 = 29°, find ∠1.
Chapter 54, Problem 30A, Use the angle values given to find 2. a. If 1 = 37, find 2. b. If 1 = 29, find 1.

Expert Solution

To determine

(a)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =56⁰.

Explanation of Solution

Given information:

∠1 =37^O

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 1

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =37^o

Now,

∠B=∠FBC+52∠B=37+52∠B=890

And

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=890∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−89−96∠D=850

Now consider a line HDGE.

∠CDE=850and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−850∠CDG=950

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG. The sum of angles is 180⁰.

∠GCD=290∠CDG=950∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−95∠2=∠CGD=560

Expert Solution

To determine

(b)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =64⁰.

Explanation of Solution

Given information:

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 2

∠1 =29^O

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =29^o

Now,

∠B=∠FBC+52∠B=29+52∠B=810

And,

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=810∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−81−96∠D=930

Now consider a line HDGE.

∠CDE=930and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−930∠CDG=870

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG, sum of angles is 180⁰_.

∠GCD=290∠CDG=870∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−87∠2=∠CGD=640

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Answer 2

Textbook Question

Chapter 54, Problem 30A

Use the angle values given to find ∠2.
a. If ∠1 = 37°, find ∠2.
b. If ∠1 = 29°, find ∠1.
Chapter 54, Problem 30A, Use the angle values given to find 2. a. If 1 = 37, find 2. b. If 1 = 29, find 1.

Expert Solution

To determine

(a)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =56⁰.

Explanation of Solution

Given information:

∠1 =37^O

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 1

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =37^o

Now,

∠B=∠FBC+52∠B=37+52∠B=890

And

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=890∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−89−96∠D=850

Now consider a line HDGE.

∠CDE=850and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−850∠CDG=950

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG. The sum of angles is 180⁰.

∠GCD=290∠CDG=950∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−95∠2=∠CGD=560

Expert Solution

To determine

(b)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =64⁰.

Explanation of Solution

Given information:

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 2

∠1 =29^O

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =29^o

Now,

∠B=∠FBC+52∠B=29+52∠B=810

And,

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=810∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−81−96∠D=930

Now consider a line HDGE.

∠CDE=930and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−930∠CDG=870

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG, sum of angles is 180⁰_.

∠GCD=290∠CDG=870∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−87∠2=∠CGD=640

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Answer 3

Textbook Question

Chapter 54, Problem 30A

Use the angle values given to find ∠2.
a. If ∠1 = 37°, find ∠2.
b. If ∠1 = 29°, find ∠1.
Chapter 54, Problem 30A, Use the angle values given to find 2. a. If 1 = 37, find 2. b. If 1 = 29, find 1.

Expert Solution

To determine

(a)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =56⁰.

Explanation of Solution

Given information:

∠1 =37^O

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 1

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =37^o

Now,

∠B=∠FBC+52∠B=37+52∠B=890

And

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=890∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−89−96∠D=850

Now consider a line HDGE.

∠CDE=850and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−850∠CDG=950

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG. The sum of angles is 180⁰.

∠GCD=290∠CDG=950∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−95∠2=∠CGD=560

Expert Solution

To determine

(b)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =64⁰.

Explanation of Solution

Given information:

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 2

∠1 =29^O

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =29^o

Now,

∠B=∠FBC+52∠B=29+52∠B=810

And,

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=810∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−81−96∠D=930

Now consider a line HDGE.

∠CDE=930and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−930∠CDG=870

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG, sum of angles is 180⁰_.

∠GCD=290∠CDG=870∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−87∠2=∠CGD=640

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Answer 4

Textbook Question

Chapter 54, Problem 30A

Use the angle values given to find ∠2.
a. If ∠1 = 37°, find ∠2.
b. If ∠1 = 29°, find ∠1.
Chapter 54, Problem 30A, Use the angle values given to find 2. a. If 1 = 37, find 2. b. If 1 = 29, find 1.

Expert Solution

To determine

(a)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =56⁰.

Explanation of Solution

Given information:

∠1 =37^O

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 1

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =37^o

Now,

∠B=∠FBC+52∠B=37+52∠B=890

And

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=890∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−89−96∠D=850

Now consider a line HDGE.

∠CDE=850and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−850∠CDG=950

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG. The sum of angles is 180⁰.

∠GCD=290∠CDG=950∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−95∠2=∠CGD=560

Expert Solution

To determine

(b)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =64⁰.

Explanation of Solution

Given information:

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 2

∠1 =29^O

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =29^o

Now,

∠B=∠FBC+52∠B=29+52∠B=810

And,

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=810∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−81−96∠D=930

Now consider a line HDGE.

∠CDE=930and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−930∠CDG=870

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG, sum of angles is 180⁰_.

∠GCD=290∠CDG=870∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−87∠2=∠CGD=640

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =56⁰.

Explanation of Solution

Given information:

∠1 =37^O

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 1

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =37^o

Now,

∠B=∠FBC+52∠B=37+52∠B=890

And

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=890∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−89−96∠D=850

Now consider a line HDGE.

∠CDE=850and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−850∠CDG=950

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG. The sum of angles is 180⁰.

∠GCD=290∠CDG=950∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−95∠2=∠CGD=560

Expert Solution

To determine

(b)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =64⁰.

Explanation of Solution

Given information:

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 2

∠1 =29^O

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =29^o

Now,

∠B=∠FBC+52∠B=29+52∠B=810

And,

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=810∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−81−96∠D=930

Now consider a line HDGE.

∠CDE=930and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−930∠CDG=870

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG, sum of angles is 180⁰_.

∠GCD=290∠CDG=870∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−87∠2=∠CGD=640

Answer 8

Expert Solution

To determine

(a)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =56⁰.

Explanation of Solution

Given information:

∠1 =37^O

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 1

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =37^o

Now,

∠B=∠FBC+52∠B=37+52∠B=890

And

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=890∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−89−96∠D=850

Now consider a line HDGE.

∠CDE=850and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−850∠CDG=950

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG. The sum of angles is 180⁰.

∠GCD=290∠CDG=950∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−95∠2=∠CGD=560

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

Value of ∠2.

Answer 13

Answer to Problem 30A

Value of ∠2 =56⁰.

Answer 14

Explanation of Solution

Given information:

∠1 =37^O

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 1

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =37^o

Now,

∠B=∠FBC+52∠B=37+52∠B=890

And

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=890∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−89−96∠D=850

Now consider a line HDGE.

∠CDE=850and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−850∠CDG=950

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG. The sum of angles is 180⁰.

∠GCD=290∠CDG=950∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−95∠2=∠CGD=560

Answer 15

Expert Solution

To determine

(b)

Value of ∠2.

Answer to Problem 30A

Value of ∠2 =64⁰.

Explanation of Solution

Given information:

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 2

∠1 =29^O

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =29^o

Now,

∠B=∠FBC+52∠B=29+52∠B=810

And,

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=810∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−81−96∠D=930

Now consider a line HDGE.

∠CDE=930and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−930∠CDG=870

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG, sum of angles is 180⁰_.

∠GCD=290∠CDG=870∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−87∠2=∠CGD=640

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

Value of ∠2.

Answer 20

Answer to Problem 30A

Value of ∠2 =64⁰.

Answer 21

Explanation of Solution

Given information:

Mathematics For Machine Technology, Chapter 54, Problem 30A , additional homework tip 2

∠1 =29^O

Calculation:

∠FBC is equal to the ∠1 because of opposite angle.

So, ∠FBC =29^o

Now,

∠B=∠FBC+52∠B=29+52∠B=810

And,

∠IAB=900and ∠IAB+∠BAF=1800So ∠BAF=1800−∠IAB∠BAF=1800−900∠BAF=900

Also,

∠HEF=840and ∠HEF+∠FED=1800So ∠FED=1800−∠HEF∠FED=1800−840∠FED=960

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

∠A=900∠B=810∠E=960So ∠D=3600−∠A−∠B−∠E∠D=360−90−81−96∠D=930

Now consider a line HDGE.

∠CDE=930and ∠CDE+∠CDG=1800So ∠CDG=1800−∠CDE∠CDG=1800−930∠CDG=870

Now consider a line BCD.

∠BCG=1510and ∠BCG+∠GCD=1800So ∠GCD=1800−∠BCG∠GCD=1800−1510∠GCD=290

Now consider a triangle CDG, sum of angles is 180⁰_.

∠GCD=290∠CDG=870∠CGD+∠GCD+∠CDG=1800∠CGD=180−∠GCD−∠CDG∠CGD=180−29−87∠2=∠CGD=640

Answer 22

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Use the angle values given to find ∠2. a. If ∠1 = 37°, find ∠2. b. If ∠1 = 29°, find ∠1.

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Answer to Problem 30A

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Answer to Problem 30A

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Chapter 54 Solutions