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


"Ken McLennan" wrote in message
...
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