trigonometry
posted by anon .
okay so i have this trigonometry problem were i HAVE to use logarithms.
Now this is my work but can u help me solve it using logarithms.
Find the area of the following triangles:(use logarithms)
1) c=426, A=45degrees 48' 36",and B=61degrees 2' 13"

C = 180  A  B = 180  45° 48' 36"  61° 2' 13" = 73° 9' 11"
Law of sines: a/sinA = c/sinC
a = c.(sinA/sinC) = 426.(sin(45° 48' 36"))/sin(73° 9' 11"))
a = 426*0.749186
a = 319.153 
i don't know why its asking the url

anyways the area would be 312,918.8

The length of a is correct.
The area of the triangle would be
A=(1/2)a*c*sin(b)
=(1/2)*319.153*426*sin(61degrees 2' 13" )
=59477.5
I am not able to reproduce your answer for the area. 
i used cosin and got that answer .. I'm i wrong .

is there anyway i can use logarthims in this problem?

In the old days when there were no calculators, surveying calculations like this would require logarithms to do the multiplications.
There was a book about an inch and a half thick called "7digit logarithm tables". We would look up sine(61° 2' 13" ) from the table, look up the logarithm of that from another table, then look up the logarithms of 426 and 319.153. These logarithms would have been added together and the antilogarithm would be the answer: 59477.5.. to 7 digits.
This is the closest relation to logarithm I can think of.
If you want, I can simulate the process.