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?
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
To solve the triangle using logarithms and half-angle formulas, it seems like you have followed the correct steps up to a certain point. However, there are some errors and inconsistencies in your calculations. Let's go through the steps and correct them:
Given parts:
a = 27
b = 21
c = 24
Using the cosine law, you correctly found the angles A, B, and C:
A = 73 deg 23' 55"
B = 48 deg 11' 23"
C = 58 deg 24' 43"
Now let's look at the outline you provided:
a = 27
b = 21
c = 25 (inconsistent with the given triangle)
To find the semiperimeter s, we use the formula: s = (a + b + c) / 2
For the given triangle, the correct value of c is 24, not 25.
Using the correct values:
a = 27
b = 21
c = 24
Now we can calculate the semiperimeter:
s = (27 + 21 + 24) / 2 = 36
Next, you correctly calculated s - a, s - b, and s - c:
s - a = 36 - 27 = 9
s - b = 36 - 21 = 15
s - c = 36 - 24 = 12
At this point, you introduce logarithms. To calculate the logarithms, you need to take the logarithm (base 10) of each value:
log(s - a) = log(9) ≈ 0.9542
log(s - b) = log(15) ≈ 1.1761
log(s - c) = log(12) ≈ 1.0792
Now, let's proceed with the rest of the steps:
1. Calculate log s using the logarithm of the semiperimeter s:
log s = log(36) ≈ 1.5563
2. Subtract log(s - a) from log s:
log r = log s - log(s - a) ≈ 1.5563 - 0.9542 ≈ 0.6021
Now, you correctly calculated log r as 0.6021.
Here's where you encountered a problem:
3. To find the value of angle A using the half-angle formula, you have:
log r = 0.6021
log(s - a) = 0.9542
However, the equation you wrote, log tan(A/2) = 9.85551 - 10, is incorrect. Instead, you should subtract log(s - a) from log r to find log tan(A/2):
log tan(A/2) = log r - log(s - a) ≈ 0.6021 - 0.9542 = -0.3521
To convert from logarithm to angle measure, we can use the antilogarithm function (10 raised to the power of the log value):
tan(A/2) ≈ 10^(-0.3521)
To find A, we can use the inverse tangent function (tan^-1) and multiply the result by 2:
A ≈ 2 * tan^(-1)(10^(-0.3521))
Similarly, you can find the values for B and C using the correct equations:
log tan(B/2) = log r - log(s - b)
log tan(C/2) = log r - log(s - c)
Once you have the values of A, B, and C from the half-angle formulas, you can check if they match the values obtained from the cosine law. The sum of the angles should be 180 degrees.
I hope this explanation helps clarify the steps you need to take to solve the triangle using logarithms and half-angle formulas. Remember to use the correct values and equations to ensure accurate results.
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 .