Looks like you are well on the right track. I haven't looked closely at your
slightly modified formulas but all your results appear same as Dana's &
mine, so I assume all OK.

Regards,
Peter T


"Cathy" wrote in message
...
I thank you both for your all this

Your formula is perfect. I just had difficulty relating your references to
A,a,B,b,C and c.

Again I realise this is probably due to a mistake in my discription. All
this trig stuff is all slowly coming back to me.

What I should have said was
Line1 = b = AC = 20° from 0° = 35 meters
Line2 = c = AB = 45° from 0° = 30 meters
Line3 = a = CB = ?° from 0° = ? Meters
(I simply forgot that side "a" would be opposite to corner "A" :)

Going by the above then
A = BAC = 45° - 20° = 25°

A=ACOS((b^2+c^2-a^2)/(2*b*c))*180/PI()
I already have A. It is 25°

or as you want internal Angle-B
B=ACOS((a^2+c^2-a^2)/(2*a*c))*180/PI()
?Should this not read:
B=ACOS((a^2+c^2-b^2)/(2*a*c))*180/PI()

Is the following therefore correct?
A°=BAC=25°
B°=ABC=ACOS((a^2+c^2-b^2)/(2*a*c))*180/PI() = 96.63893451
C°=ACB=ACOS((b^2+a^2-c^2)/(2*a*b))*180/PI() = 58.36374786

And modelling by previous spreadsheet
# A B C
1 Side Direction Length
2 Line1 20 35
3 Line2 45 30
4 Line3 ? ?

The formula for B4 is:
=180-(ACOS((C4^2+C2^2-C3^2)/(2*C4*C2))*180/PI())+B2
= 141.6362521°

and Formula for C4 is:
=ROUND(SQRT(C2^2+C3^2-2*C2*C3*COS((B3-B2)*PI()/180)),2)
=14.89139507 meters

I have just realised another mistake in my original example, i.e. the
angle
of c should have been in the opposite direction. I should be able to work
that out myself and correct what I am trying to achieve.

Thank you so much for all your help.

Kind regards
C