Question

Use logarithms and the law of tangents to solve the triangle ABC, given that a=21.46 ft, b=46.28 ft, and C=32°28'30"

Using the previous values,
a=21.46
b=46.28
C=32-28-30=32.475
We have
A+B=180-32.475=147.525
a+b=67.74
a-b=-24.82

tan((A+B)/2)=3.433633
tan((A-B)/2)=tan((A+B)/2)*(a-b)/(a+b)
= -1.2580863 <<.. i did this part in logarithms it was actually very easy now that i understood it. thank you!

but i cant seem to understand what u did in the following part .

(A-B)/2 = atan(1.2580863)= -51.520285
A-B = -103.04057

I didn't get what u did there and when i tried it i came up with different numbers. Can you please clarify the steps.

thank you for taking the time to help me. I really appreciate it.

A=(147.525-103.04057)/2=22.242215°
B=(147.525+103.04057)/2=125.282785°

Answer 1

Sorry, it must have been a typo.
It should read:
(A-B)/2 = atan(-1.2580863)= -51.520285
So multiply by 2 to get
A-B = -103.04057

The negative sign was because I did not choose A and B wisely. It turned out that A<B.

So if
A-B=-103.04057....(1)
A+B=147.525.....(2)
Add (1) and (2)
2A+0B=44.484430 => A=22.242215
Subtract (1) from (2)
0A+2B=250.56557 => B=125.282785

Notice that B>A, so A-B is negative, so is arcTan((A-B)/2).