Geometry For Enjoyment And Challenge
Geometry For Enjoyment And Challenge
91st Edition
ISBN: 9780866099653
Author: Richard Rhoad, George Milauskas, Robert Whipple
Publisher: McDougal Littell
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Question
Chapter 2.4, Problem 1PSA

a)

To determine

To Find: 3

a)

Expert Solution
Check Mark

Answer to Problem 1PSA

  49

Explanation of Solution

Given information: 2 is complementary to 3 . 4=131

Formula used: linear pair axiom, properties of intersecting lines and angles formed between the lines.

2 and 3 are complementary, hence 2+3=90....(1)

Also, by hence pair axiom, 3+4=180 .

Given 4=131 , hence 3=181131=49

b)

To determine

To Find: 6

b)

Expert Solution
Check Mark

Answer to Problem 1PSA

  131

Explanation of Solution

Given information: 2 is complementary to 3 . 4=131

Formula used: linear pair axiom, properties of intersecting lines and angles formed between the lines.

  6=131 (Vertically opposite angles)

c)

To determine

To Find: 5

c)

Expert Solution
Check Mark

Answer to Problem 1PSA

  49

Explanation of Solution

Given information: 2 is complementary to 3 . 4=131

Formula used: linear pair axiom, properties of intersecting lines and angles formed between the lines.

  5=49 (Vertically opposite angles)

3rd

  2+3=90 (Complementary angles)

d)

To determine

To Find: 2

d)

Expert Solution
Check Mark

Answer to Problem 1PSA

  41

Explanation of Solution

Given information: 2 is complementary to 3 . 4=131

Formula used: linear pair axiom, properties of intersecting lines and angles formed between the lines.

Put 3=49 we get 2=41

e)

To determine

To Find: 1

e)

Expert Solution
Check Mark

Answer to Problem 1PSA

  139

Explanation of Solution

Given information: 2 is complementary to 3 . 4=131

Formula used: linear pair axiom, properties of intersecting lines and angles formed between the lines.

  1+2=180 (Linear pair)

  1+41=180

  1=139

f)

To determine

To Find: 8

f)

Expert Solution
Check Mark

Answer to Problem 1PSA

  41

Explanation of Solution

Given information: 2 is complementary to 3 . 4=131

Formula used: linear pair axiom, properties of intersecting lines and angles formed between the lines.

  2=8 (Vertically opposite angles)

  8=41

g)

To determine

To Find: 7

g)

Expert Solution
Check Mark

Answer to Problem 1PSA

  139

Explanation of Solution

Given information: 2 is complementary to 3 . 4=131

Formula used: linear pair axiom, properties of intersecting lines and angles formed between the lines.

  7=1 (Vertically opposite angle)

  7=139

Chapter 2 Solutions

Geometry For Enjoyment And Challenge

Ch. 2.1 - Prob. 11PSBCh. 2.1 - Prob. 12PSBCh. 2.1 - Prob. 13PSBCh. 2.1 - Prob. 14PSCCh. 2.1 - Prob. 15PSCCh. 2.2 - Prob. 1PSACh. 2.2 - Prob. 2PSACh. 2.2 - Prob. 3PSACh. 2.2 - Prob. 4PSACh. 2.2 - Prob. 5PSACh. 2.2 - Prob. 6PSACh. 2.2 - Prob. 7PSACh. 2.2 - Prob. 8PSACh. 2.2 - Prob. 9PSACh. 2.2 - Prob. 10PSACh. 2.2 - Prob. 11PSACh. 2.2 - Prob. 12PSBCh. 2.2 - Prob. 13PSBCh. 2.2 - Prob. 14PSBCh. 2.2 - Prob. 15PSBCh. 2.2 - Prob. 16PSBCh. 2.2 - Prob. 17PSBCh. 2.2 - Prob. 18PSBCh. 2.2 - Prob. 19PSBCh. 2.2 - Prob. 20PSBCh. 2.2 - Prob. 21PSBCh. 2.2 - Prob. 22PSBCh. 2.2 - Prob. 23PSBCh. 2.2 - Prob. 24PSBCh. 2.2 - Prob. 25PSCCh. 2.2 - Prob. 26PSCCh. 2.3 - Prob. 1PSACh. 2.3 - Prob. 2PSACh. 2.3 - Prob. 3PSACh. 2.3 - Prob. 4PSACh. 2.3 - Prob. 5PSACh. 2.3 - Prob. 6PSACh. 2.3 - Prob. 7PSACh. 2.3 - Prob. 8PSACh. 2.3 - Prob. 9PSBCh. 2.3 - Prob. 10PSBCh. 2.3 - Prob. 11PSBCh. 2.3 - Prob. 12PSBCh. 2.3 - Prob. 13PSCCh. 2.3 - Prob. 14PSCCh. 2.4 - Prob. 1PSACh. 2.4 - Prob. 2PSACh. 2.4 - Prob. 3PSACh. 2.4 - Prob. 4PSACh. 2.4 - Prob. 5PSACh. 2.4 - Prob. 6PSACh. 2.4 - Prob. 7PSACh. 2.4 - Prob. 8PSACh. 2.4 - Prob. 9PSACh. 2.4 - Prob. 10PSBCh. 2.4 - Prob. 11PSBCh. 2.4 - Prob. 12PSBCh. 2.4 - Prob. 13PSBCh. 2.4 - Prob. 14PSBCh. 2.4 - Prob. 15PSBCh. 2.4 - Prob. 16PSBCh. 2.4 - Prob. 17PSBCh. 2.4 - Prob. 18PSBCh. 2.4 - Prob. 19PSBCh. 2.4 - Prob. 20PSCCh. 2.4 - Prob. 21PSCCh. 2.5 - Prob. 1PSACh. 2.5 - Prob. 2PSACh. 2.5 - Prob. 3PSACh. 2.5 - Prob. 4PSACh. 2.5 - Prob. 5PSACh. 2.5 - Prob. 6PSACh. 2.5 - Prob. 7PSACh. 2.5 - Prob. 8PSACh. 2.5 - Prob. 9PSACh. 2.5 - Prob. 10PSACh. 2.5 - Prob. 11PSBCh. 2.5 - Prob. 12PSBCh. 2.5 - Prob. 13PSBCh. 2.5 - Prob. 14PSBCh. 2.5 - Prob. 15PSBCh. 2.5 - Prob. 16PSBCh. 2.5 - Prob. 17PSCCh. 2.5 - Prob. 18PSCCh. 2.5 - Prob. 19PSCCh. 2.6 - Prob. 1PSACh. 2.6 - Prob. 2PSACh. 2.6 - Prob. 3PSACh. 2.6 - Prob. 4PSACh. 2.6 - Prob. 5PSACh. 2.6 - Prob. 6PSACh. 2.6 - Prob. 7PSACh. 2.6 - Prob. 8PSACh. 2.6 - Prob. 9PSACh. 2.6 - Prob. 10PSACh. 2.6 - Prob. 11PSBCh. 2.6 - Prob. 12PSBCh. 2.6 - Prob. 13PSBCh. 2.6 - Prob. 14PSBCh. 2.6 - Prob. 15PSCCh. 2.6 - Prob. 16PSCCh. 2.7 - Prob. 1PSACh. 2.7 - Prob. 2PSACh. 2.7 - Prob. 3PSACh. 2.7 - Prob. 4PSACh. 2.7 - Prob. 5PSACh. 2.7 - Prob. 6PSACh. 2.7 - Prob. 7PSACh. 2.7 - Prob. 8PSACh. 2.7 - Prob. 9PSACh. 2.7 - Prob. 10PSBCh. 2.7 - Prob. 11PSBCh. 2.7 - Prob. 12PSBCh. 2.7 - Prob. 13PSBCh. 2.7 - Prob. 14PSBCh. 2.7 - Prob. 15PSBCh. 2.7 - Prob. 16PSCCh. 2.7 - Prob. 17PSCCh. 2.7 - Prob. 18PSDCh. 2.7 - Prob. 19PSDCh. 2.7 - Prob. 20PSDCh. 2.8 - Prob. 1PSACh. 2.8 - Prob. 2PSACh. 2.8 - Prob. 3PSACh. 2.8 - Prob. 4PSACh. 2.8 - Prob. 5PSACh. 2.8 - Prob. 6PSACh. 2.8 - Prob. 7PSACh. 2.8 - Prob. 8PSACh. 2.8 - Prob. 9PSBCh. 2.8 - Prob. 10PSBCh. 2.8 - Prob. 11PSBCh. 2.8 - Prob. 12PSBCh. 2.8 - Prob. 13PSBCh. 2.8 - Prob. 14PSBCh. 2.8 - Prob. 15PSCCh. 2 - Prob. 1RPCh. 2 - Prob. 2RPCh. 2 - Prob. 3RPCh. 2 - Prob. 4RPCh. 2 - Prob. 5RPCh. 2 - Prob. 6RPCh. 2 - Prob. 7RPCh. 2 - Prob. 8RPCh. 2 - Prob. 9RPCh. 2 - Prob. 10RPCh. 2 - Prob. 11RPCh. 2 - Prob. 12RPCh. 2 - Prob. 13RPCh. 2 - Prob. 14RPCh. 2 - Prob. 15RPCh. 2 - Prob. 16RPCh. 2 - Prob. 17RPCh. 2 - Prob. 18RPCh. 2 - Prob. 19RPCh. 2 - Prob. 20RPCh. 2 - Prob. 21RPCh. 2 - Prob. 22RPCh. 2 - Prob. 23RPCh. 2 - Prob. 24RPCh. 2 - Prob. 25RPCh. 2 - Prob. 26RPCh. 2 - Prob. 27RPCh. 2 - Prob. 28RPCh. 2 - Prob. 29RPCh. 2 - Prob. 30RPCh. 2 - Prob. 31RPCh. 2 - Prob. 32RPCh. 2 - Prob. 33RPCh. 2 - Prob. 34RPCh. 2 - Prob. 35RPCh. 2 - Prob. 36RPCh. 2 - Prob. 37RPCh. 2 - Prob. 38RP

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