Posted by **Anon** on Sunday, February 27, 2011 at 2:11am.

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°

- trigonometry (repost) Mathmate -
**MathMate**, Sunday, February 27, 2011 at 7:21am
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).

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