Geometry, Trig, Calculus . . . not sure

[Mei_Mienzi]

Hello:

I am not sure if my question belongs on this forum or not, but either way I am hoping that I can get some help. I have attached a picture of my problem, hoping that it will help (I'm a visual person). You will note a rectangle representing a piece of wood that needs to be cut as noted in red. Adjacent to the rectangle is a right triangle for which the base and hypotenuse are unknown. I know that by extending the base of the rectangle and the line above it (not sure what you call it) that the two lines will meet, forming the right triangle. However, I do not know how far either of those line have to be extended to the point of intersection. If I knew that I could calculate the desired angle of cut. I would think that this would be a common problem for a carpenter, which I am not, and therefore not all that hard to figure out (though I haven't a clue). Can someone help me? Thanks.

 

[stapel]

I am not sure if my question belongs on this forum or not, but either way I am hoping that I can get some help.

The category should probably correspond to the course you're taking. At a guess, that'll be trig.

I have attached a picture of my problem, hoping that it will help (I'm a visual person).

block and triangle:

*-------------+-------*
|             |       |
+ .           |       |
|    '  .     |       |
|1"        '  .       |
|        0.75"|  '  . |          
*-------------+-------@
|<--39.875"-->|<--x-->|

You will note a rectangle representing a piece of wood that needs to be cut as noted in red. Adjacent to the rectangle is a right triangle for which the base and hypotenuse are unknown. I know that by extending the base of the rectangle and the line above it (not sure what you call it) that the two lines will meet, forming the right triangle. However, I do not know how far either of those line have to be extended to the point of intersection. If I knew that I could calculate the desired angle of cut. I would think that this would be a common problem for a carpenter, which I am not, and therefore not all that hard to figure out (though I haven't a clue). Can someone help me? Thanks.

Since you're wanting an angle, I'd set up the trig ratio and solve for the angle. You've got the angle, which I've denoted as "@" above, in the triangle. You have base and height info for the left-hand polygon. Try setting up similar ratios. Since the "opposite over adjacent" is the tangent ratio (here), you can set up as follows, using similar triangles:

. . . . .\(\displaystyle \dfrac{1}{0.75\, +\, x}\, =\, \dfrac{0.75}{x}\, =\, \tan(@)\)

You can solve the first two "halves" for the value of "x" (here and (here)), and then back-solve for the value of the angle (here).

If you get stuck, please reply with a clear listing of your thoughts and efforts so far (keeping in mind that the above leads to an "ideal" mathematical solution, suited to your math class; a "real world" solution might require a professional contractor). Thank you!