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it says using a ruler and a protractor @mathstudent55
in my hand :/
Just follow this video. https://www.youtube.com/watch?v=t4xCOUNEInI
I only see "Draw a 105 degree angle" above.
would you like me to re write the question ? @mathstudent55
Here's my way of doing it.
Start with a line, and mark two points on the line. |dw:1440958648016:dw|
okay ive done that
We are going to draw a square segment AB as a side. |dw:1440958758007:dw|
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|
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|
Now open the compass more than AD, center is at C and center it at D and draw two arcs above A.
Connect A to E and extend. |dw:1440959255726:dw|
FE is perpendicular to AB. Also, FA = AB
Now open the compass to AB. Center the compass at F and draw and arc to intersect the arc above B. |dw:1440959563117:dw|
G is the 4th vertex of our square.
We have square ABGF. Since it's a square, all vertex angles are 90 degrees. |dw:1440959604880:dw|
Now we extend line BG up above point G. |dw:1440959701330:dw|
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|
GH is half of BG.
Open the compass to GH. Center the compass at G and mark a point above G on line BG. |dw:1440960034307:dw|
Sorry. I realize now I made a mistake when I tried to figure this out earlier. We don't need half the side.
Let's go back to the square.
Now we draw a diagonal in the square.
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|
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.
A 60-deg angle is not difficult to construct because it is the measure of the vertex angles of an equilateral triangle.
Now we construct an equilteral triangle at the upper left vertex of the square.
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|
Angle PFG measures 60 deg. Angle GFB measures 45 deg. Angle PFB measures 105 deg.
With the help of a square, its diagonal, and an equilateral triangle, we now have a 105-deg angle.