(a) Answer m ∠ 2 = 90 ∘ , m ∠ 3 = 90 ∘ and m ∠ 4 = 90 ∘ Explanation Given information: Draw a line A B → . Choose a point O between A and B. The protractor is used. m ∠ 1 = 90 ∘ Formula used: Angle addition postulate Calculation: Consider the Angle addition postulate When m ∠ 1 = 90 ∘ From the above figure it can be observe that a → and b → both represents straight angles . The following equations are: ∠ 1 + ∠ 2 = 180 ∘ ............ ( 1 ) ∠ 1 + ∠ 4 = 180 ∘ ............ ( 2 ) ∠ 3 + ∠ 4 = 180 ∘ ............ ( 3 ) Therefore, from equation (1), (2) and (3) 90 ∘ + ∠ 2 = 180 ∘ ∠ 2 = 180 ∘ − 90 ∘ ∠ 2 = 90 ∘ ∠ 1 + ∠ 4 = 180 ∘ 90 ∘ + ∠ 4 = 180 ∘ ∠ 4 = 180 ∘ − 90 ∘ ∠ 4 = 90 ∘ And ∠ 3 + ∠ 4 = 180 ∘ ∠ 3 + 90 ∘ = 180 ∘ ∠ 3 = 180 ∘ − 90 ∘ ∠ 3 = 90 ∘ Therefore, m ∠ 2 = 90 ∘ , m ∠ 3 = 90 ∘ and m ∠ 4 = 90 ∘ Conclusion: m ∠ 2 = 90 ∘ , m ∠ 3 = 90 ∘ and m ∠ 4 = 90 ∘ (b) To determine To calculate: The value of m ∠ 2 , m ∠ 3 and m ∠ 4 Answer m ∠ 2 = 87 ∘ , m ∠ 3 = 93 ∘ and m ∠ 4 = 87 ∘ Explanation Given information: Draw a line A B → . Choose a point O between A and B. The protractor is used. m ∠ 1 = 93 ∘ Formula used: Angle addition postulate Calculation: Consider the Angle addition postulate When m ∠ 1 = 93 ∘ As it can be seen from the above figure it can be observe that a → and b → both represents straight angles. The following equations are: ∠ 1 + ∠ 2 = 180 ∘ ............ ( 1 ) ∠ 1 + ∠ 4 = 180 ∘ ............ ( 2 ) ∠ 3 + ∠ 4 = 180 ∘ ............ ( 3 ) Therefore, from equation (1), (2) and (3) 93 ∘ + ∠ 2 = 180 ∘ ∠ 2 = 180 ∘ − 93 ∘ ∠ 2 = 87 ∘ ∠ 1 + ∠ 4 = 180 ∘ 93 ∘ + ∠ 4 = 180 ∘ ∠ 4 = 180 ∘ − 93 ∘ ∠ 4 = 87 ∘ And ∠ 3 + ∠ 4 = 180 ∘ ∠ 3 + 87 ∘ = 180 ∘ ∠ 3 = 180 ∘ − 87 ∘ ∠ 3 = 93 ∘ Therefore, m ∠ 2 = 87 ∘ , m ∠ 3 = 93 ∘ and m ∠ 4 = 87 ∘ Conclusion: m ∠ 2 = 87 ∘ , m ∠ 3 = 93 ∘ and m ∠ 4 = 87 ∘

McDougal Littell Jurgensen Geometry: Student Edition Geometry

5th Edition

ISBN: 9780395977279

Author: Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Publisher: Houghton Mifflin Company College Division

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Question

Chapter 1.4, Problem 26WE

(a)

To determine

To find: The value of m∠2, m∠3 and m∠4

(a)

Expert Solution

Answer to Problem 26WE

m∠2=90∘, m∠3=90∘ and m∠4=90∘

Explanation of Solution

Given information:

Draw a line AB→ . Choose a point O between A and B. The protractor is used. m∠1=90∘

Formula used:

Angle addition postulate

Calculation:

Consider the Angle addition postulate

When m∠1=90∘

From the above figure it can be observe that a→ and b→ both represents straight angles.

The following equations are:

∠1+∠2=180∘............(1)∠1+∠4=180∘............(2)∠3+∠4=180∘............(3)

Therefore, from equation (1), (2) and (3)

90∘+∠2=180∘ ∠2=180∘−90∘ ∠2=90∘

∠1+∠4=180∘90∘+∠4=180∘ ∠4=180∘−90∘ ∠4=90∘

And

∠3+∠4=180∘∠3+90∘=180∘ ∠3=180∘−90∘ ∠3=90∘

Therefore,

m∠2=90∘, m∠3=90∘ and m∠4=90∘

Conclusion:

m∠2=90∘, m∠3=90∘ and m∠4=90∘

(b)

To determine

To calculate: The value of m∠2, m∠3 and m∠4

(b)

Expert Solution

Answer to Problem 26WE

m∠2=87∘, m∠3=93∘ and m∠4=87∘

Explanation of Solution

Given information:

Draw a line AB→ . Choose a point O between A and B. The protractor is used. m∠1=93∘

Formula used:

Angle addition postulate

Calculation:

Consider the Angle addition postulate

When m∠1=93∘

As it can be seen from the above figure it can be observe that a→ and b→ both represents straight angles.

The following equations are:

∠1+∠2=180∘............(1)∠1+∠4=180∘............(2)∠3+∠4=180∘............(3)

Therefore, from equation (1), (2) and (3)

93∘+∠2=180∘ ∠2=180∘−93∘ ∠2=87∘

∠1+∠4=180∘93∘+∠4=180∘ ∠4=180∘−93∘ ∠4=87∘

And

∠3+∠4=180∘∠3+87∘=180∘ ∠3=180∘−87∘ ∠3=93∘

Therefore,

m∠2=87∘, m∠3=93∘ and m∠4=87∘

Conclusion:

m∠2=87∘, m∠3=93∘ and m∠4=87∘

Chapter 1.4, Problem 25WE

Chapter 1.4, Problem 27WE

Chapter 1 Solutions

McDougal Littell Jurgensen Geometry: Student Edition Geometry

Ch. 1.1 - Prob. 1CE Ch. 1.1 - Prob. 2CE Ch. 1.1 - Prob. 3CE Ch. 1.1 - Prob. 4CE Ch. 1.1 - Prob. 5CE Ch. 1.1 - Prob. 6CE Ch. 1.1 - Prob. 7CE Ch. 1.1 - Prob. 8CE Ch. 1.1 - Prob. 1WE Ch. 1.1 - Prob. 2WE

Ch. 1.1 - Prob. 3WE Ch. 1.1 - Prob. 4WE Ch. 1.1 - Prob. 5WE Ch. 1.1 - Prob. 6WE Ch. 1.1 - Prob. 7WE Ch. 1.1 - Prob. 8WE Ch. 1.1 - Prob. 9WE Ch. 1.1 - Prob. 10WE Ch. 1.2 - Prob. 1CE Ch. 1.2 - Prob. 2CE Ch. 1.2 - Prob. 3CE Ch. 1.2 - Prob. 4CE Ch. 1.2 - Prob. 5CE Ch. 1.2 - Prob. 6CE Ch. 1.2 - Prob. 7CE Ch. 1.2 - Prob. 8CE Ch. 1.2 - Prob. 9CE Ch. 1.2 - Prob. 10CE Ch. 1.2 - Prob. 11CE Ch. 1.2 - Prob. 12CE Ch. 1.2 - Prob. 13CE Ch. 1.2 - Prob. 14CE Ch. 1.2 - Prob. 15CE Ch. 1.2 - Prob. 16CE Ch. 1.2 - Prob. 17CE Ch. 1.2 - Prob. 18CE Ch. 1.2 - Prob. 19CE Ch. 1.2 - Prob. 20CE Ch. 1.2 - Prob. 1WE Ch. 1.2 - Prob. 2WE Ch. 1.2 - Prob. 3WE Ch. 1.2 - Prob. 4WE Ch. 1.2 - Prob. 5WE Ch. 1.2 - Prob. 6WE Ch. 1.2 - Prob. 7WE Ch. 1.2 - Prob. 8WE Ch. 1.2 - Prob. 9WE Ch. 1.2 - Prob. 10WE Ch. 1.2 - Prob. 11WE Ch. 1.2 - Prob. 12WE Ch. 1.2 - Prob. 13WE Ch. 1.2 - Prob. 14WE Ch. 1.2 - Prob. 15WE Ch. 1.2 - Prob. 16WE Ch. 1.2 - Prob. 17WE Ch. 1.2 - Prob. 18WE Ch. 1.2 - Prob. 19WE Ch. 1.2 - Prob. 20WE Ch. 1.2 - Prob. 21WE Ch. 1.2 - 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