Solving for Angle C in Triangle ABC | Napier's Analogy Explained

[zorro]

Homework Statement

In a triangle ABC, a=6 b=3 cos(A-B)=4/5. Find the angle C.

Homework Equations

The Attempt at a Solution

here we need to find tan(A-B/2)
I used the formula tan2x=2tanx/(1/tan^2x)
and got 2 values of tan(A-B/2) as -3 and 1/3
On what explanation do I reject one of them?
-90<A-B/2 <90
so tanA-B/2 can be both positive and negative.
Please explain in detail

 

[LCKurtz]

Using the following formula with $\theta = A - B$

$$\tan\frac \theta 2 = \frac {1-\cos(\theta)}{\sin(\theta)}$$

I get $A - B = \pm 1/3$

Check what this gives for C using Napier's identity and I think your question will be answered.

 

[zorro]

I assume u mean tan(A-B/2) = +/- 1/3
How did the negative sign come? You took two values for sin(theta) ?
After solving I got c= -90 or c=90
both can be correct

 

[LCKurtz]

I assume u mean tan(A-B/2) = +/- 1/3

Yes.

How did the negative sign come? You took two values for sin(theta) ?

Yes

After solving I got c= -90 or c=90
both can be correct

A triangle with -90 degrees? I don't think so. And you can check, using the fact that it is a right triangle, that the numbers all work.

 

[zorro]

Thanks