Draw line AB, with A and B 2" apart. Then construct....

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Cinnamon

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Sorry to bother you guys again but I need help with one more if possible........


Draw Line AB with points A and B 2 inches apart. Construct a line perpendicular to AB at point A. Locate point X on the perpendicular 1 inch from point A (to the south) and then draw line BX. Number the 4 angles with vertex A as angles 1-4, numbering counterclockwise starting with angle BAX as angle 1. Number the four angles with vertex B as angles 5-8 numbering counter clockwise starting with angle ABX as angle 5. Number the four angles with vertex X as angles 9-12 numbering counterclockwise starting with angle AXB as angle 9.

I know it's alot and my own best friends mom who's a teacher didnt even get it please help me if u can
 
Re: I need help with One More

Hello, Cinnamon!

Draw Line AB with points A and B 2 inches apart.
Construct a line perpendicular to AB at point A.
Locate point X on the perpendicular 1 inch from point A (to the south) and then draw line BX.

Number the four angles with vertex A as angles 1-4, numbering CCW starting with angle BAX as angle 1.
Number the four angles with vertex B as angles 5-8 numbering CCW starting with angle ABX as angle 5.
Number the four angles with vertex X as angles 9-12 numbering CCW starting with angle AXB as angle 9.
Click to expand...
Code: 
      |                 /
    3 | 2           8 / 7
 - - A* - - - - - - *B - -
    4 | 1       5 / 6
      |         /
      |       /
      |     /
      | 9 /
   10 | /
     X* 12
    / |
  / 11|
We're told that \(\displaystyle AB\,=\,2,\;AX\,=\,1\)

Okay, we have the diagram . . . What's the question ??
 
WELLL

now i have the measure in degrees for angles 1-4.....i need 2 know how to measure the rest of them
 
Re: WELLL

Hello, Cinnamon!

Well, we already know that: \(\displaystyle \,\angle1\,=\,\angle2 \,=\,\angle3\,=\,\angle 4\,=\,90^o\) . . . right?

If we can find one of the other angles, we can determine all the angles.


We're told that: \(\displaystyle \,AB\,=\,2,\;AC\,=\,1\)

So: \(\displaystyle \,\tan(\angle 9)\,=\,\frac{2}{1}\;\;\Rightarrow\;\;\angle9\,\approx\,63.4^o\;\;\Rightarrow\;\;\angle11\,=\,63.4^o\)

Then: \(\displaystyle \,\angle10\:=\:180^o\,-\,63.4^o\:=\:116.6^o \:=\:\angle 12\)


Since \(\displaystyle \angle5\) is the other acute angle in right triangle \(\displaystyle ABC\),

\(\displaystyle \;\;\angle 5\;=\;90^o\,-\,63.4^o\;=\;26.6^o\;=\;\angle 7\)

Finally: \(\displaystyle \,\angle 6\;=\;180^o\,-\,26.6^o\;=\;153.4^o\;=\;\angle 8\)
 
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