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 35WE

(a)

To determine

To find: The number of rays and angles.

(a)

Expert Solution

Answer to Problem 35WE

The figure (c) has 6 angles and figure (d) has 10 angles.

Explanation of Solution

Given information:

The given figure can be named as:

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 1.4, Problem 35WE , additional homework tip 1

Formula used:

Property of Non collinear rays.

Calculation:

As it can be seen from the above figure it can be observe that figure (c) has 4 rays and 6 angle that is,

∠AOC,∠AOB,∠AOD,∠COB,∠COD and ∠BOD

Similarly figure (d) has 5 rays and 10 angles that is,

∠AOE,∠AOC,∠AOB,∠AOD,∠EOC,∠EOB,∠EOD,∠COB,∠COD and ∠BOD

Therefore, the figure (c) has 6 angles and figure (d) has 10 angles.

Conclusion:

The figure (c) has 6 angles and figure (d) has 10 angles.

(b)

To determine

To get: The number of angles formed by six non collinear rays that have the same end point.

(b)

Expert Solution

Answer to Problem 35WE

It can be predicted that six non collinear rays having same end point will have 15 angles.

Explanation of Solution

Given information:

The given figure can be named as:

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 1.4, Problem 35WE , additional homework tip 2

Formula used:

Property of Non collinear rays.

Calculation:

As it can be seen from the above figure it can be observe that angles are increasing in a particular pattern with the increase in no of rays.

From the first to last figure as rays are increasing with the difference of one, the angles are increasing with difference of 2, 3 and 4.

Therefore, it can be predicted that six non collinear rays having same end point will have 15 angles.

Conclusion:

It can be predicted that six non collinear rays having same end point will have 15 angles.

(c)

To determine

To find:The expression that gives the number of angles formed by nnon collinear rays that have the same endpoint.

(c)

Expert Solution

Answer to Problem 35WE

The general expression for the number of angles formed by n rays having same endpoint is n(n−1)2

Explanation of Solution

Given information:

The given figure can be named as:

McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 1.4, Problem 35WE , additional homework tip 3

Formula used:

Non collinear rays is used

Calculation:

From the above figure it can be observe that the angles are increasing in a particular order.

Thus the general expression for the number of angles formed by n rays having same endpoint is n(n−1)2

This can be verified by substituting various values such as 1, 2, ….

Conclusion:

The general expression for the number of angles formed by n rays having same endpoint is n(n−1)2

Chapter 1.4, Problem 34WE

Chapter 1.4, Problem 36WE

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 - Prob. 22WE Ch. 1.2 - Prob. 23WE Ch. 1.2 - Prob. 24WE Ch. 1.2 - Prob. 25WE Ch. 1.2 - Prob. 26WE Ch. 1.2 - Prob. 27WE Ch. 1.2 - Prob. 28WE Ch. 1.2 - Prob. 29WE Ch. 1.2 - Prob. 30WE Ch. 1.2 - Prob. 31WE Ch. 1.2 - Prob. 32WE Ch. 1.2 - Prob. 33WE Ch. 1.2 - Prob. 34WE Ch. 1.2 - Prob. 35WE Ch. 1.2 - Prob. 36WE Ch. 1.2 - Prob. 1ST1 Ch. 1.2 - Prob. 2ST1 Ch. 1.2 - Prob. 3ST1 Ch. 1.2 - Prob. 4ST1 Ch. 1.2 - Prob. 5ST1 Ch. 1.2 - Prob. 6ST1 Ch. 1.2 - Prob. 7ST1 Ch. 1.2 - Prob. 1AR Ch. 1.2 - Prob. 2AR Ch. 1.2 - Prob. 3AR Ch. 1.2 - Prob. 4AR Ch. 1.2 - Prob. 5AR Ch. 1.2 - Prob. 6AR Ch. 1.2 - Prob. 7AR Ch. 1.2 - Prob. 8AR Ch. 1.2 - Prob. 9AR Ch. 1.2 - Prob. 10AR Ch. 1.2 - Prob. 11AR Ch. 1.2 - Prob. 12AR Ch. 1.2 - Prob. 13AR Ch. 1.2 - Prob. 14AR Ch. 1.2 - Prob. 15AR Ch. 1.2 - Prob. 16AR Ch. 1.2 - Prob. 17AR Ch. 1.2 - Prob. 18AR Ch. 1.2 - Prob. 19AR Ch. 1.2 - Prob. 20AR Ch. 1.2 - Prob. 21AR Ch. 1.2 - Prob. 22AR Ch. 1.2 - Prob. 23AR Ch. 1.2 - Prob. 24AR Ch. 1.2 - Prob. 25AR Ch. 1.2 - Prob. 26AR Ch. 1.2 - Prob. 27AR Ch. 1.2 - Prob. 28AR Ch. 1.2 - Prob. 29AR Ch. 1.2 - Prob. 30AR Ch. 1.2 - Prob. 31AR Ch. 1.2 - Prob. 32AR Ch. 1.2 - Prob. 33AR Ch. 1.2 - Prob. 34AR Ch. 1.2 - Prob. 35AR Ch. 1.2 - Prob. 36AR Ch. 1.2 - Prob. 37AR Ch. 1.2 - Prob. 38AR Ch. 1.2 - Prob. 39AR Ch. 1.2 - Prob. 40AR Ch. 1.2 - Prob. 41AR Ch. 1.2 - Prob. 42AR Ch. 1.2 - Prob. 43AR Ch. 1.2 - Prob. 44AR Ch. 1.2 - Prob. 45AR Ch. 1.2 - Prob. 46AR Ch. 1.3 - Prob. 1CE Ch. 1.3 - Prob. 2CE Ch. 1.3 - Prob. 3CE Ch. 1.3 - Prob. 4CE Ch. 1.3 - Prob. 5CE Ch. 1.3 - Prob. 6CE Ch. 1.3 - Prob. 7CE Ch. 1.3 - Prob. 8CE Ch. 1.3 - Prob. 9CE Ch. 1.3 - Prob. 10CE Ch. 1.3 - Prob. 11CE Ch. 1.3 - Prob. 12CE Ch. 1.3 - Prob. 13CE Ch. 1.3 - Prob. 14CE Ch. 1.3 - Prob. 15CE Ch. 1.3 - Prob. 16CE Ch. 1.3 - Prob. 17CE Ch. 1.3 - Prob. 18CE Ch. 1.3 - Prob. 19CE Ch. 1.3 - Prob. 20CE Ch. 1.3 - Prob. 21CE Ch. 1.3 - Prob. 22CE Ch. 1.3 - Prob. 23CE Ch. 1.3 - Prob. 24CE Ch. 1.3 - Prob. 25CE Ch. 1.3 - Prob. 26CE Ch. 1.3 - Prob. 1WE Ch. 1.3 - Prob. 2WE Ch. 1.3 - Prob. 3WE Ch. 1.3 - Prob. 4WE Ch. 1.3 - Prob. 5WE Ch. 1.3 - Prob. 6WE Ch. 1.3 - Prob. 7WE Ch. 1.3 - Prob. 8WE Ch. 1.3 - Prob. 9WE Ch. 1.3 - Prob. 10WE Ch. 1.3 - Prob. 11WE Ch. 1.3 - Prob. 12WE Ch. 1.3 - Prob. 13WE Ch. 1.3 - Prob. 14WE Ch. 1.3 - Prob. 15WE Ch. 1.3 - Prob. 16WE Ch. 1.3 - Prob. 17WE Ch. 1.3 - Prob. 18WE Ch. 1.3 - Prob. 19WE Ch. 1.3 - Prob. 20WE Ch. 1.3 - Prob. 21WE Ch. 1.3 - Prob. 22WE Ch. 1.3 - Prob. 23WE Ch. 1.3 - Prob. 24WE Ch. 1.3 - Prob. 25WE Ch. 1.3 - Prob. 26WE Ch. 1.3 - Prob. 27WE Ch. 1.3 - Prob. 28WE Ch. 1.3 - Prob. 29WE Ch. 1.3 - Prob. 30WE Ch. 1.3 - Prob. 31WE Ch. 1.3 - Prob. 32WE Ch. 1.3 - Prob. 33WE Ch. 1.3 - Prob. 34WE Ch. 1.3 - Prob. 35WE Ch. 1.3 - Prob. 36WE Ch. 1.3 - Prob. 37WE Ch. 1.3 - Prob. 38WE Ch. 1.3 - Prob. 39WE Ch. 1.3 - Prob. 40WE Ch. 1.3 - Prob. 41WE Ch. 1.3 - Prob. 42WE Ch. 1.3 - Prob. 43WE Ch. 1.3 - Prob. 44WE Ch. 1.3 - Prob. 45WE Ch. 1.3 - Prob. 46WE Ch. 1.3 - Prob. 47WE Ch. 1.3 - Prob. 48WE Ch. 1.4 - Prob. 1CE Ch. 1.4 - Prob. 2CE Ch. 1.4 - Prob. 3CE Ch. 1.4 - Prob. 4CE Ch. 1.4 - Prob. 5CE Ch. 1.4 - Prob. 6CE Ch. 1.4 - Prob. 7CE Ch. 1.4 - Prob. 8CE Ch. 1.4 - Prob. 9CE Ch. 1.4 - Prob. 10CE Ch. 1.4 - Prob. 11CE Ch. 1.4 - Prob. 12CE Ch. 1.4 - Prob. 13CE Ch. 1.4 - Prob. 14CE Ch. 1.4 - Prob. 15CE Ch. 1.4 - Prob. 16CE Ch. 1.4 - Prob. 17CE Ch. 1.4 - Prob. 18CE Ch. 1.4 - Prob. 19CE Ch. 1.4 - Prob. 20CE Ch. 1.4 - Prob. 21CE Ch. 1.4 - Prob. 22CE Ch. 1.4 - Prob. 23CE Ch. 1.4 - Prob. 24CE Ch. 1.4 - Prob. 25CE Ch. 1.4 - Prob. 26CE Ch. 1.4 - Prob. 27CE Ch. 1.4 - Prob. 28CE Ch. 1.4 - Prob. 29CE Ch. 1.4 - Prob. 30CE Ch. 1.4 - Prob. 31CE Ch. 1.4 - Prob. 32CE Ch. 1.4 - Prob. 33CE Ch. 1.4 - Prob. 34CE Ch. 1.4 - Prob. 35CE Ch. 1.4 - Prob. 36CE Ch. 1.4 - Prob. 37CE Ch. 1.4 - Prob. 38CE Ch. 1.4 - Prob. 1WE Ch. 1.4 - Prob. 2WE Ch. 1.4 - Prob. 3WE Ch. 1.4 - Prob. 4WE Ch. 1.4 - Prob. 5WE Ch. 1.4 - Prob. 6WE Ch. 1.4 - Prob. 7WE Ch. 1.4 - Prob. 8WE Ch. 1.4 - Prob. 9WE Ch. 1.4 - Prob. 10WE Ch. 1.4 - Prob. 11WE Ch. 1.4 - Prob. 12WE Ch. 1.4 - Prob. 13WE Ch. 1.4 - Prob. 14WE Ch. 1.4 - Prob. 15WE Ch. 1.4 - Prob. 16WE Ch. 1.4 - Prob. 17WE Ch. 1.4 - Prob. 18WE Ch. 1.4 - Prob. 19WE Ch. 1.4 - Prob. 20WE Ch. 1.4 - Prob. 21WE Ch. 1.4 - Prob. 22WE Ch. 1.4 - Prob. 23WE Ch. 1.4 - Prob. 24WE Ch. 1.4 - Prob. 25WE Ch. 1.4 - Prob. 26WE Ch. 1.4 - Prob. 27WE Ch. 1.4 - Prob. 28WE Ch. 1.4 - Prob. 29WE Ch. 1.4 - Prob. 30WE Ch. 1.4 - Prob. 31WE Ch. 1.4 - Prob. 32WE Ch. 1.4 - Prob. 33WE Ch. 1.4 - Prob. 34WE Ch. 1.4 - Prob. 35WE Ch. 1.4 - Prob. 36WE Ch. 1.5 - Prob. 1CE Ch. 1.5 - Prob. 2CE Ch. 1.5 - Prob. 3CE Ch. 1.5 - Prob. 4CE Ch. 1.5 - Prob. 5CE Ch. 1.5 - Prob. 6CE Ch. 1.5 - Prob. 7CE Ch. 1.5 - Prob. 8CE Ch. 1.5 - Prob. 9CE Ch. 1.5 - Prob. 10CE Ch. 1.5 - Prob. 11CE Ch. 1.5 - Prob. 12CE Ch. 1.5 - Prob. 13CE Ch. 1.5 - Prob. 14CE Ch. 1.5 - Prob. 15CE Ch. 1.5 - Prob. 16CE Ch. 1.5 - Prob. 1WE Ch. 1.5 - Prob. 2WE Ch. 1.5 - Prob. 3WE Ch. 1.5 - Prob. 4WE Ch. 1.5 - Prob. 5WE Ch. 1.5 - Prob. 6WE Ch. 1.5 - Prob. 7WE Ch. 1.5 - Prob. 8WE Ch. 1.5 - Prob. 9WE Ch. 1.5 - Prob. 10WE Ch. 1.5 - Prob. 11WE Ch. 1.5 - Prob. 12WE Ch. 1.5 - Prob. 13WE Ch. 1.5 - Prob. 14WE Ch. 1.5 - Prob. 15WE Ch. 1.5 - Prob. 16WE Ch. 1.5 - Prob. 17WE Ch. 1.5 - Prob. 18WE Ch. 1.5 - Prob. 19WE Ch. 1.5 - Prob. 20WE Ch. 1.5 - Prob. 1E Ch. 1.5 - Prob. 2E Ch. 1.5 - Prob. 3E Ch. 1.5 - Prob. 4E Ch. 1.5 - Prob. 5E Ch. 1.5 - Prob. 6E Ch. 1.5 - Prob. 1ST2 Ch. 1.5 - Prob. 2ST2 Ch. 1.5 - Prob. 3ST2 Ch. 1.5 - Prob. 4ST2 Ch. 1.5 - Prob. 5ST2 Ch. 1.5 - Prob. 6ST2 Ch. 1.5 - Prob. 7ST2 Ch. 1.5 - Prob. 8ST2 Ch. 1.5 - Prob. 9ST2 Ch. 1.5 - Prob. 10ST2 Ch. 1.5 - Prob. 11ST2 Ch. 1.5 - Prob. 12ST2 Ch. 1 - Prob. 1CR Ch. 1 - Prob. 2CR Ch. 1 - Prob. 3CR Ch. 1 - Prob. 4CR Ch. 1 - Prob. 5CR Ch. 1 - Prob. 6CR Ch. 1 - Prob. 7CR Ch. 1 - Prob. 8CR Ch. 1 - Prob. 9CR Ch. 1 - Prob. 10CR Ch. 1 - Prob. 11CR Ch. 1 - Prob. 12CR Ch. 1 - Prob. 13CR Ch. 1 - Prob. 14CR Ch. 1 - Prob. 15CR Ch. 1 - Prob. 16CR Ch. 1 - Prob. 17CR Ch. 1 - Prob. 18CR Ch. 1 - Prob. 19CR Ch. 1 - Prob. 20CR Ch. 1 - Prob. 1CT Ch. 1 - Prob. 2CT Ch. 1 - Prob. 3CT Ch. 1 - Prob. 4CT Ch. 1 - Prob. 5CT Ch. 1 - Prob. 6CT Ch. 1 - Prob. 7CT Ch. 1 - Prob. 8CT Ch. 1 - Prob. 9CT Ch. 1 - Prob. 10CT Ch. 1 - Prob. 11CT Ch. 1 - Prob. 12CT Ch. 1 - Prob. 13CT Ch. 1 - Prob. 14CT Ch. 1 - Prob. 15CT Ch. 1 - Prob. 16CT Ch. 1 - Prob. 17CT Ch. 1 - Prob. 18CT Ch. 1 - Prob. 19CT Ch. 1 - Prob. 20CT Ch. 1 - Prob. 21CT Ch. 1 - Prob. 22CT Ch. 1 - Prob. 23CT Ch. 1 - Prob. 24CT

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The number of rays and angles .

Videos

To find: The number of rays and angles.

Answer to Problem 35WE

Explanation of Solution

To get: The number of angles formed by six non collinear rays that have the same end point.

Answer to Problem 35WE

Explanation of Solution

To find:The expression that gives the number of angles formed by nnon collinear rays that have the same endpoint.

Answer to Problem 35WE

Explanation of Solution

Chapter 1 Solutions

Additional Math Textbook Solutions