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anonymous

  • one year ago

Draw a 105 degree angle

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  1. mathstudent55
    • one year ago
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    |dw:1440958077956:dw|

  2. anonymous
    • one year ago
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    it says using a ruler and a protractor @mathstudent55

  3. anonymous
    • one year ago
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    in my hand :/

  4. Jhannybean
    • one year ago
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    Just follow this video. https://www.youtube.com/watch?v=t4xCOUNEInI

  5. mathstudent55
    • one year ago
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    I only see "Draw a 105 degree angle" above.

  6. anonymous
    • one year ago
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    would you like me to re write the question ? @mathstudent55

  7. mathstudent55
    • one year ago
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    No need.

  8. mathstudent55
    • one year ago
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    Here's my way of doing it.

  9. mathstudent55
    • one year ago
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    Start with a line, and mark two points on the line. |dw:1440958648016:dw|

  10. anonymous
    • one year ago
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    okay ive done that

  11. mathstudent55
    • one year ago
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    We are going to draw a square segment AB as a side. |dw:1440958758007:dw|

  12. mathstudent55
    • one year ago
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    Open the compass to AB and center it at A and draw an arc. Using the same opening, center it at B and draw an arc. |dw:1440958818783:dw|

  13. mathstudent55
    • one year ago
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    Now we want to draw a perpendicular to line AB through point A and a perpendicular to line AB through point B. Open the compass to any opening between AB. Center the compass at A and draw and two arcs on line AB, one on each side of A. |dw:1440959091893:dw|

  14. mathstudent55
    • one year ago
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    |dw:1440959116221:dw|

  15. mathstudent55
    • one year ago
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    Now open the compass more than AD, center is at C and center it at D and draw two arcs above A.

  16. mathstudent55
    • one year ago
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    |dw:1440959214600:dw|

  17. mathstudent55
    • one year ago
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    Connect A to E and extend. |dw:1440959255726:dw|

  18. mathstudent55
    • one year ago
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    FE is perpendicular to AB. Also, FA = AB

  19. mathstudent55
    • one year ago
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    Now open the compass to AB. Center the compass at F and draw and arc to intersect the arc above B. |dw:1440959563117:dw|

  20. mathstudent55
    • one year ago
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    G is the 4th vertex of our square.

  21. mathstudent55
    • one year ago
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    We have square ABGF. Since it's a square, all vertex angles are 90 degrees. |dw:1440959604880:dw|

  22. mathstudent55
    • one year ago
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    |dw:1440959655179:dw|

  23. mathstudent55
    • one year ago
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    Now we extend line BG up above point G. |dw:1440959701330:dw|

  24. mathstudent55
    • one year ago
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    Now we need half the length of the segment BG, so we draw its perpendicular bisector. Open the compass to more than half of BG. Center the compass at G and draw two arcs. Then center the compass at B and draw two arcs that intersect the previous two arcs. Connect the intersections of the arcs with a segment. This segment is the perpendicular bisector of segment BG. |dw:1440959917480:dw|

  25. mathstudent55
    • one year ago
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    GH is half of BG.

  26. mathstudent55
    • one year ago
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    Open the compass to GH. Center the compass at G and mark a point above G on line BG. |dw:1440960034307:dw|

  27. mathstudent55
    • one year ago
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    Sorry. I realize now I made a mistake when I tried to figure this out earlier. We don't need half the side.

  28. mathstudent55
    • one year ago
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    Let's go back to the square.

  29. mathstudent55
    • one year ago
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    |dw:1440960509048:dw|

  30. mathstudent55
    • one year ago
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    Now we draw a diagonal in the square.

  31. mathstudent55
    • one year ago
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    Since we constructed a square, we know each vertex angle measures 90 degrees. Also, when we draw a diagonal, that bisects the vertex angle, so we have now two 45 degree angles at the vertex. |dw:1440960561813:dw|

  32. mathstudent55
    • one year ago
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    We notice that 105 degrees = 45 degrees + 60 degrees. If we place a 60 degree angle at that vertex, then 60 deg + 45 deg = 105 deg, which is what we want.

  33. mathstudent55
    • one year ago
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    A 60-deg angle is not difficult to construct because it is the measure of the vertex angles of an equilateral triangle.

  34. mathstudent55
    • one year ago
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    Now we construct an equilteral triangle at the upper left vertex of the square.

  35. mathstudent55
    • one year ago
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    |dw:1440960839669:dw|

  36. mathstudent55
    • one year ago
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    Open the compass to AB and center at F. Draw and arc. Then center at G and draw an arc. Where the arcs intersect, it's the third vertex of equilateral triangle FGP. |dw:1440960981941:dw|

  37. mathstudent55
    • one year ago
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    |dw:1440961010150:dw|

  38. mathstudent55
    • one year ago
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    Angle PFG measures 60 deg. Angle GFB measures 45 deg. Angle PFB measures 105 deg.

  39. mathstudent55
    • one year ago
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    With the help of a square, its diagonal, and an equilateral triangle, we now have a 105-deg angle.

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