Trig Function seems odd.

[Ken McLennan]

G'day there One & All,

I've used XL to try to solve a simple trig problem, but the
results have thrown me somewhat. I've no doubt the fault is mine, but I
can see what's wrong.

The problem is to determine timber thicknesses along a right
angled taper:

+--5--+--5--+--5--+--5--+--5--+
|
5
|
+


I want to determine the distance from each '+' to the hypotenuse.
It should be simple enough. I use the formula '=1/TAN(5/25)' to
determine the angle, (4.933154876) and then use '=A11*TAN(RADIANS($B
$8))'. [$B$8 is where the angle is stored. A11:A15 holds the series
5,10,15,20,25].

I get the series 0.4, 0.9, 1.3, 1.7, 2.2 as a result. I know this
is incorrect since the problem itself gives the final member as 5.

I think I may have neglected a conversion between degrees &
radians somewhere, but when I try to resolve it that way the results are
more ridiculous than what I have already. Can anyone there see what I've
done wrong? The web searches I've done so far either confirm the
formulae as correct, or explain the functions in such a complex manner
that I don't understand what's going on (you can probably guess that my
high school trig knowledge is just a trifle rusty).

Thanks for listening,
Ken McLennan
Qld, Australia

 

[Ken McLennan]

The problem is to determine timber thicknesses along a right
angled taper:

+--5--+--5--+--5--+--5--+--5--+
|
5
|
+

I've just noticed in my post that the diagram is a bit askew.
It's supposed to be a simple right triangle.

See ya,
Ken McLennan
Qld, Australia

 

[Rick Rothstein \(MVP - VB\)]

This problem does not require Trig at all... you can do it with simple
proportions. Since this sounds like homework, I'll let you set up the
solution on your own. However, draw the hypotenuse, and then drop the
perpendiculars from the + signs down to the hypotenuse. The ratio between
any two sides (think the two sides that are not hypotenuses) in each smaller
triangle formed by dropping the perpendiculars is equal to the ratio between
the same two sides in the large triangle. Since you know two sides in the
large triangle and you know one side those sides in each of the smaller
triangles, you can calculate the missing side.

Rick

 

[Rick Rothstein \(MVP - VB\)]

It is if you copy it and drop it into Notepad (where the font is a
fixed-width one instead of the proportional one used by default in
newsreaders).

Rick

 

[Ken McLennan]

G'day there Rick,

It is if you copy it and drop it into Notepad (where the font is a
fixed-width one instead of the proportional one used by default in
newsreaders).

You know, even after several years of using Gravity, I never even
noticed it used a proportional font!! Kinda says something about my
powers of observation!

See ya
Thanks for your help
Ken

 

[Ken McLennan]

G'day there Rick,

This problem does not require Trig at all... you can do it with simple
proportions. Since this sounds like homework, I'll let you set up the
solution on your own.

Chuckle!!! If I was still young enough to be doing school

However, draw the hypotenuse, and then drop the
perpendiculars from the + signs down to the hypotenuse. The ratio between
any two sides (think the two sides that are not hypotenuses) in each smaller
triangle formed by dropping the perpendiculars is equal to the ratio between
the same two sides in the large triangle. Since you know two sides in the
large triangle and you know one side those sides in each of the smaller
triangles, you can calculate the missing side.

Aaaahhh yes... our old friend Pythagorus. I think he took
sick and died a few years back.

It's rather obvious when you explain it to me in one syllable
words

You're quite right of course, and just as I was reading your post
I realised that it's also a simple matter of proportion - each 5 units
is 20% of the adjacent side and so the opposite side can be divided into
20% sections to determine the results.

I think the real problem was that when I tried to be smart and
used XL I couldn't get it to work right and that became the problem I
concentrated on; my triangles became a side issue.

So, I can now easily determine the length of the sides, but why
is my spreadsheet giving incorrect results? What have I gotten wrong
there?

See ya
Thanks for your reply.
Ken McLennan
Qld, Australia

 

[David Biddulph]

=1/TAN(5/25) doesn't give an angle, but =ATAN(5/25) does

And having calculated the angle that way (already in radians), you don't
need the RADIANS function before using that in the TAN function.

 

[Ken McLennan]

G'day there David,

=1/TAN(5/25) doesn't give an angle, but =ATAN(5/25) does

And having calculated the angle that way (already in radians), you don't
need the RADIANS function before using that in the TAN function.

AHA!!!! It's so much easier when you know what you're doing!

Thanks very much for that. I thought that ATAN was an Arctangent
and had some other relationship to the angle and that I needed the
inverse of the tangent. Oh well. Maybe I should have tried to remember
my school stuff. Or learned it to begin with

I've changed the formulae and all is now well with the world.
However I think it would be much better if Microsoft built a spreadsheet
program that just knew what I wanted and gave me the answers without me
having to accurately formulate the questions

Thanks once again for your help.
Ken McLennan
Qld, Australia.

 

[David Biddulph]

ATAN is indeed the arctangent, and that's what you were trying to calculate.

I agree with your point that the difficulty is with knowing what question to
ask, rather than in getting Excel to provide the answer.

Perhaps your thought of using the inverse of the tangent is a consequence of
being in the antipodes?

 

[Ken McLennan]

G'day there David,

ATAN is indeed the arctangent, and that's what you were trying to calculate.

Gotta get back to the books I s'pose. Or just blunder on in
ignorance. The latter has served me OK for the last 30 years or so

I agree with your point that the difficulty is with knowing what question to
ask, rather than in getting Excel to provide the answer.

As they say "A little knowledge is a dangerous thing", and my
knowledge of trigonometry is quite little.

Perhaps your thought of using the inverse of the tangent is a consequence of
being in the antipodes?

Hey!! There's an idea. I hadn't thought of that. After all, the
sun does go anticlockwise down here. Widdershins instead of deosil <g>.

See ya,
Thanks again,
Ken