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This is an outline that i have to use to find the solution of the question. It involves logarithms and half-angle formulas.I ha...Asked by Anon
This is an outline that i have to use to find the solution of the question. It involves logarithms and half-angle 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
s-a=9.5
s-b=15
s-c=11.5
log(s-a)= .97772
log(s-b)=1.19033
log(s-c)=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(s-a) = 0.97772
--------
log tan A/2 = 9.85551 - 10
A = ? << can you help me convert.
log r = 0.83323
log (s-b) = 1.19033
--------
log tan B/2 = 9.64290 - 10
B = ? same here my brain freezes
log r = 0.83323
log (s-c) = 1.06070
--------
log tan C/2 = 9.77253 - 10
C = ?
A+B+C= 180 deg 0' 2"
do they match?
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
s-a=9.5
s-b=15
s-c=11.5
log(s-a)= .97772
log(s-b)=1.19033
log(s-c)=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(s-a) = 0.97772
--------
log tan A/2 = 9.85551 - 10
A = ? << can you help me convert.
log r = 0.83323
log (s-b) = 1.19033
--------
log tan B/2 = 9.64290 - 10
B = ? same here my brain freezes
log r = 0.83323
log (s-c) = 1.06070
--------
log tan C/2 = 9.77253 - 10
C = ?
A+B+C= 180 deg 0' 2"
do they match?
Answers
Answered by
Reiny
I noticed an error in your outline
log(s-b) = 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
log(s-b) = 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
Answered by
Anon
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 .
=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 .
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