Math Help

[CloudyIFR]

I'm working on a forumula for MS Excel Spreadsheet.

I know the length along the bottom (3.598) and top (2.156) and the angle of the leading edge (21 degrees), what would the formula be to find the length along the right vertical edge? I put it into a CAD to get the result of 3.766 but I can't figure out the formula to find that length to use in Excel.

Thanks so much for the help.
Curtis
Montana

[slipstick]

You can either do it using simple trig
x = 3.766 / Cos (21 degrees)

or using basic Pythagoras theorem
x = Sqr (3.766^2 + (3.598 - 2.156)^2

Either way the answer comes out as about 4.03

Steve

[Derfy]

How about trying this:

=(3.598-2.156)/TAN(RADIANS(21))

which, with any luck, should give you this:

3.756538431

Anyway, it does in my version of Excel. :D

:cool:

[JetPlaneFlyer]

You can either do it using simple trig
x = 3.766 / Cos (21 degrees)

or using basic Pythagoras theorem
x = Sqr (3.766^2 + (3.598 - 2.156)^2

Either way the answer comes out as about 4.03

Steve

I think that you just worked out the length of the hypotenuse rather than the adjacent ;)

Steve

[adam_one]

Try this:

Right vertical edge = (3.598-2.156)*Tan (69 deg)

:)

[slipstick]

I think that you just worked out the length of the hypotenuse rather than the adjacent ;)

Steve
You're right....didn't read carefully enough. Somehow it just didn't occur to me that he was actually looking for the length of the side with the length marked on it :(.

Steve

[CloudyIFR]

Thanks guys, I really appreciate it.

I have one more that I can't figure out.

I'm looking for the Excel formula for the distance ??? in the photo. I know the the angle and the two distances but need the ??? distance in the photo.

Thanks
Curtis

[BarrelRoll]

Thanks guys, I really appreciate it.

I have one more that I can't figure out.

I'm looking for the Excel formula for the distance ??? in the photo. I know the the angle and the two distances but need the ??? distance in the photo.

Thanks
Curtis

In case it's not a joke, the ??? distance is 3.6 :)

[Murocflyer]

In case it's not a joke, the ??? distance is 3.6 :)

Show your work please. :)

Frank

[CloudyIFR]

Wow, did I screw that up or what! :-)

Here's what I'm actually looking for.

I've input what I know and I'm looking for the distance.

I realize it looks like it'd be = 3.6 - 2.16 but the angle must be 30 degrees making the entire leading/trailing edge sweep back futher than shown.

Thanks

PS I hadn't had my second cup of coffee yet!

[Murocflyer]

...

[CloudyIFR]

I'm looking for the Excel formula to find the distance in green. The CAD program shows the value should be 3.517.

Curtis
Montana

[MarkusN]

Is 6 the vertical distance? Then it's easy. ??? = 6 / sqrt(3)

(tan (30°) is 1/sqrt(3))


Edit: OK, you gave the additional info. It definitely is 6 * tan (alpha).

[CloudyIFR]

PERFECT!!!

Thanks so much.
Curtis
Montana