Expert: Paul Klarreich - 5/2/2008

Question
can you give it a try
use the logarithms and the law of tangents to solve the triangle ABC,given that a=21.46ft,b=46.28ft,c=32°28'30".


The sides given are a and b, with angle C, which is the angle between them.  The law of cos() states that c^2 = a^2 + b^2 - 2*a*b*cos(c).  Using this, angles a and b could be found as well by shifting the variables to make c be a different included angle between 2 sides.
Answer     (a - b)/(a + b) = (tan((A - B)/2)))/(tan((A + B)/2)))



A + B + C = 180
A + B = 180 - C
A + B = 180 - 32°28'30" or 147°2'00"

I will need to rearrange it so that it will work out.

in this way i'm using a to mean b and b to mean a, or else i will have a negative log.

tan((A + B)/2) = tan((147°2'00")/2) or tan(73°31'00")

log((46.28 - 21.46)/(46.28 + 21.46)) = log(24.82/67.74)

log(tan((A - B)/2)/tan((A + B)/2))) =
log(tan((A - B)/2))) - log(tan((A + B)/2)) =
log(tan((A - B)/2))) - log(tan(73°31'00"))

log(tan((A - B)/2))) - log(tan(73°31'00")) = log(24.82/67.74)
log(tan((A - B)/2))) = log(24.82/67.74) + log(tan(73°31'00")))
log(tan((A - B)/2))) = log((24.72/67.74)tan(73°31'00")))
tan((A - B)/2) = (24.72/67.24)tan(73°31'00")

(A - B)/2 = tan^-1((24.72/67.24)tan(73°31'00"))
A - B = 2tan^-1((24.72/67.24)tan(73°31'00"))
A = 2tan^-1((24.72/67.24)tan(73°31'00")) + B

A + B =
2tan^-1((24.72/67.24)tan(73°31'00")) + B + B = 147°2'00"
2B = 44°41'30.04
B = 22°20'45.02

A = 2tan^-1((24.72/67.24)tan(73°31'00")) + 22°20'45.02
A = 125°25'31.16

Keeping in mind that i had to switch the true values of a and b to b and a, so therefore the actual angles for the correct one i switched back.

B = 125°25'31.16
A = 22°20'45.02
C = 32°28'30"

Proof:

c^2 = 21.46^2 + 46.28^2 - 2(21.46 * 46.28)cos(32°28'30")
c = 30.59

21.46/sin(A) = 30.59/sin(32°28'30")
sin(A) =

no matter what i still can't get it to work.


Answer
Questioner:   grace
Category:  Advanced Math
Private:  No
 
Subject:  plane trigonometry
Question:  can you give it a try
use the logarithms and the law of tangents to solve the triangle ABC,given that a=21.46ft, b=46.28ft, C=32°28'30".
.................................
Hi, Grace,

You wrote (more):

.................
The sides given are a and b, with angle C, which is the angle between them.  The law of cos() states that c^2 = a^2 + b^2 - 2*a*b*cos(c).  Using this, angles a and b could be found as well by shifting the variables to make c be a different included angle between 2 sides.

Answer   (a - b)/(a + b) = (tan((A - B)/2)))/(tan((A + B)/2)))

A + B + C = 180
A + B = 180 - C
A + B = 180 - 32°28'30" or 147°2'00"

>> Wait a minute.  That's  147-32-30  [I can't make those degree signs, sorry.]

I will need to rearrange it so that it will work out.

>> I am going to copy your stuff and make changes, OK? [OF course it's OK, what are you going to do about it?]

in this way i'm using a to mean b and b to mean a, or else i will have a negative log.

tan((A + B)/2) = tan((147°2'00")/2) or tan(73°31'00")

log((46.28 - 21.46)/(46.28 + 21.46)) = log(24.82/67.74)

log(tan((A - B)/2)/tan((A + B)/2))) =
log(tan((A - B)/2))) - log(tan((A + B)/2)) =
log(tan((A - B)/2))) - log(tan(73°31'00"))

log(tan((A - B)/2))) - log(tan(73°31'00")) = log(24.82/67.74)
log(tan((A - B)/2))) = log(24.82/67.74) + log(tan(73°31'00")))
log(tan((A - B)/2))) = log((24.72/67.74)tan(73°31'00")))
tan((A - B)/2) = (24.72/67.24)tan(73°31'00")

(A - B)/2 = tan^-1((24.72/67.24)tan(73°31'00"))
A - B = 2tan^-1((24.72/67.24)tan(73°31'00"))
A = 2tan^-1((24.72/67.24)tan(73°31'00")) + B

A + B =
2tan^-1((24.72/67.24)tan(73°31'00")) + B + B = 147°2'00"
2B = 44°41'30.04
B = 22°20'45.02

A = 2tan^-1((24.72/67.24)tan(73°31'00")) + 22°20'45.02
A = 125°25'31.16

Keeping in mind that i had to switch the true values of a and b to b and a, so therefore the actual angles for the correct one i switched back.

B = 125°25'31.16
A = 22°20'45.02
C = 32°28'30"

Proof:

c^2 = 21.46^2 + 46.28^2 - 2(21.46 * 46.28)cos(32°28'30")
c = 30.59

21.46/sin(A) = 30.59/sin(32°28'30")
sin(A) =

no matter what i still can't get it to work.
-----------------COPIED HERE-----------------------------------
A + B = 180 - 32°28'30" or 147°32'30"

tan((A + B)/2) = tan((147°32'30")/2) or tan(73°46'15")

Now use (a - b)/(a + b) = (tan((A - B)/2)))/(tan((A + B)/2)))

OR CHANGE THAT TO:

(b - a)/(b + a) = (tan((B - A)/2)))/(tan((B + A)/2)))

so you don't have to switch a's and b's.

a=21.46ft, b=46.28ft, C=32°28'30".

Let's write the thing out first:

(46.28 - 21.46)/(46.28 + 21.46)) = tan((B - A)/2))/tan(73°46'15")
(24.82)/(67.74) = tan((B - A)/2))/tan(73°46'15")

log((46.28 - 21.46)/(46.28 + 21.46)) = log(24.82/67.74)

>> Yes.

Now write:

tan(73°46'15") (24.82)/(67.74) = tan((B - A)/2))

I think you just want to compute this, first.  YOu are making real heavy weather of it [too complicated].

log tan((B - A)/2)) = log tan(73°46'15") + log 24.82 -  log 67.74

Now, if you don't mind, I'm going to use Excel for this. (The Windows calculator is simplest, but you did say......)

[Note that all YOUR measurements are degrees, but Excel likes radians, so I/we have to do pi/180 to get radians, and 180/pi to get back degrees.

We also need  deg-min-sec to deg.xxx first: deg.xxx = deg + min/60 + sec/3600.]

log tan((B - A)/2)) = log tan(73°46'15") + log 24.82 -  log 67.74

log tan((B - A)/2)) = 0.535994058 + 1.394801777 - 1.830845192

log tan((B - A)/2)) = 0.099950643

tan((B - A)/2)) = 1.258782345

(B - A)/2  = 51.53560016 degrees  << Excel likes ATAN(1.25..*180/PI)

B - A = 103.0712003 degrees.

B + A = 147.5416667 degrees.

B = 125.3064335 degrees.

A = 22.23523318

>> NOT BAD -- your answers are pretty close to this, so I think you did fine up to this point.

log(tan((A - B)/2)/tan((A + B)/2))) =
log(tan((A - B)/2))) - log(tan((A + B)/2)) =
log(tan((A - B)/2))) - log(tan(73°31'00"))

log(tan((A - B)/2))) - log(tan(73°31'00")) = log(24.82/67.74)
log(tan((A - B)/2))) = log(24.82/67.74) + log(tan(73°31'00")))
log(tan((A - B)/2))) = log((24.72/67.74)tan(73°31'00")))
tan((A - B)/2) = (24.72/67.24)tan(73°31'00")

(A - B)/2 = tan^-1((24.72/67.24)tan(73°31'00"))
A - B = 2tan^-1((24.72/67.24)tan(73°31'00"))
A = 2tan^-1((24.72/67.24)tan(73°31'00")) + B

A + B =
2tan^-1((24.72/67.24)tan(73°31'00")) + B + B = 147°2'00"
2B = 44°41'30.04  
B = 22°20'45.02

>> YES, THAT LOOKS RIGHT.

A = 2tan^-1((24.72/67.24)tan(73°31'00")) + 22°20'45.02
A = 125°25'31.16

Keeping in mind that i had to switch the true values of a and b to b and a, so therefore the actual angles for the correct one i switched back.

B = 125°25'31.16
A = 22°20'45.02
C = 32°28'30"

Proof:

c^2 = 21.46^2 + 46.28^2 - 2(21.46 * 46.28)cos(32°28'30")
c = 30.59

>> Now I think this is 'cheating'.  You are using the L of cosines. Perfectly correct, of course, but you are supposed to use the L of tangents. (but who's counting?)

21.46/sin(A) = 30.59/sin(32°28'30")
sin(A) =

What's the problem?  You have, so far:

a = 21.46ft,
b = 46.28ft,
c = 30.59

A = 22°20'45.02
B = 125°25'31.16
C = 32°28'30".

i.e you have everything.

You want to check the law of sines?

21.46/sin(22°20'45.02) = 30.59/sin(32°28'30")

Looks OK to me.  After reading through this, you deserve a good nights sleep.  Say Goodnight, Gracie.