Geometry For Enjoyment And Challenge
Geometry For Enjoyment And Challenge
91st Edition
ISBN: 9780866099653
Author: Richard Rhoad, George Milauskas, Robert Whipple
Publisher: McDougal Littell
Question
Book Icon
Chapter 2, Problem 36RP

a

To determine

To Find : The coordinates of B for given condition.

a

Expert Solution
Check Mark

Answer to Problem 36RP

   The coordinates of B = (- 1 , - 2 )

Explanation of Solution

Given: We are given that with respect to Origin, point A=(1,2) is rotated 100° clockwise, then 80° counterclockwise, then 210° clockwise ,and finally 50° counterclockwise to point B.

  Geometry For Enjoyment And Challenge, Chapter 2, Problem 36RP , additional homework tip  1

Concept Used:

Using rule of clockwise and counterclockwise rotation where clockwise stands for negative and counterclockwise stands for positive movement of angles.

Calculation:

As shown in the figure the point O is showing origin and A is representing the location of coordinates (1, 2 ).Also we have been given the movement of point A clockwise and counterclockwise

Applying the rule of clockwise and counterclockwise rotation on given information we can break down the given information and write following points:

  1) 100° clockwise = - 100°2) 80° counterclockwise = + 80°3) 210° clockwise = - 210°4) 50° counterclockwise = +50°

As we are given after all these movements point B is obtained. Thus, to know the position of point B adding all the movements:

  = - 100 ° + 80° - 210° + 50°= -180°

Thus, point B is at the clockwise rotation of 180° from A. Which is shown by following graph.

  Geometry For Enjoyment And Challenge, Chapter 2, Problem 36RP , additional homework tip  2

So the coordinates of point B will be ( - 1, -2 ) as it is at 180° clockwise rotation from A.

Conclusion:

The coordinates of B = (- 1 , - 2 )

b

To determine

To Find: The rotation after which the point is in first quadrant again.

b

Expert Solution
Check Mark

Answer to Problem 36RP

The point was in first quadrant after second rotation.

Explanation of Solution

Given: We are given that with respect to Origin, point A=(1,2) is rotated 100° clockwise, then 80° counterclockwise, then 210° clockwise ,and finally 50° counterclockwise to point B.

  Geometry For Enjoyment And Challenge, Chapter 2, Problem 36RP , additional homework tip  3

Concept Used:

Using rule of clockwise and counterclockwise rotation where clockwise stands for negative and counterclockwise stands for positive movement of angles.

Calculation:

As shown in the figure the point O is showing origin and A is representing the location of coordinates (1, 2 ).Also we have been given the movement of point A clockwise and counterclockwise

Now as the point A is in first quadrant to know that when again it will come in same quadrant applying the rule of clockwise and counterclockwise rotation on given information, we can break down the given information and write following points:

  1) 100° clockwise = - 100°

After this movement the point will move in the fourth quadrant.

  2) 80° counterclockwise = + 80°

After this movement the point will come back in the first quadrant.

After this for next two rotation it will shift to third quadrant and stay there.

So, after second rotation the point again will come in first quadrant

Conclusion:

The point was in first quadrant after second rotation

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
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Elementary Geometry For College Students, 7e
Geometry
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Cengage,
Text book image
Elementary Geometry for College Students
Geometry
ISBN:9781285195698
Author:Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:Cengage Learning