trigonometry
posted by Anon .
This is an outline that i have to use to find the solution of the question. It involves logarithms and halfangle formulas.I have worked though it but got stuck on some parts. I'd like if you look over my work.
solve the triangle for which given parts are
a=27 ,b=21 ,c=24
Now i used the cosine law and got the answers
A=73 deg 23' 55"
B=48 deg 11' 23"
C=58 deg 24' 43"
...................
okay now for this outline;
a=27
b=21
c=25

2s=73
s=36.5
sa=9.5
sb=15
sc=11.5
log(sa)= .97772
log(sb)=1.19033
log(sc)=1.06070

=3.22875
log s =1.56229

2)1.66646

log r =0.83323
now this part I'm having trouble with.
log r = 0.83323
log(sa) = 0.97772

log tan A/2 = 9.85551  10
A = ? << can you help me convert.
log r = 0.83323
log (sb) = 1.19033

log tan B/2 = 9.64290  10
B = ? same here my brain freezes
log r = 0.83323
log (sc) = 1.06070

log tan C/2 = 9.77253  10
C = ?
A+B+C= 180 deg 0' 2"
do they match?

I noticed an error in your outline
log(sb) = log 15 = 1.17609 you have 1.19033
Your outline suggests to me Heron's formula for finding the area of a triangle.
I didn't see anything about area of triangle in your question
then I see log r =0.83323
where does r come from ?
To do a calculation like
log tan A/2 = 9.85551  10
A = ?
I had already shown you how to do that in one of your previous posts
http://www.jiskha.com/display.cgi?id=1298862691 
that was a typo its 15.5 which makes the log entry correct.
=3.22875
log s =1.56229 subtract.

2)1.66646 <<.. then divide by 2 =

log r =0.83323
the reason why i posted the out line was because they did not match. and i didn't understand how you got the inverse tangent .