# Two angles and the angle in between in 180 degrees; maximization

• Jun 14th 2011, 06:01 AM
Zellator
Two angles and the angle in between in 180 degrees; maximization
Hi Forum!
I have a question

We are to find the x coordinate of P to maximize abp
a(1,2) [use your imagination here, the graph is not right hahaha]
b(-1,1)
c(-1,0)
d(1,0)

Now, if we consider t as the x coordinate of P, we have:
Tan(A)=2/1-t
Tan(B)=1/1+t

If we have to find Tan(C), how should we do it?
A+B+C=180
C=180-A-B
Where do we go from there?

We know that tan(180) is 0, but how are we going to use it?

Thanks.
• Jun 14th 2011, 11:51 AM
Opalg
Re: Two angles and the angle in between in 180 degrees; maximization
Quote:

Originally Posted by Zellator
Hi Forum!
I have a question

We are to find the x coordinate of P to maximize abp
a(1,2) [use your imagination here, the graph is not right hahaha]
b(-1,1)
c(-1,0)
d(1,0)

Now, if we consider t as the x coordinate of P, we have:
Tan(A)=2/1-t
Tan(B)=1/1+t

If we have to find Tan(C), how should we do it?
A+B+C=180
C=180-A-B
Where do we go from there?

We know that tan(180) is 0, but how are we going to use it?

Thanks.

$\tan(180^\circ-C) = -\tan C$, and so $\tan C = -\tan(A+B) = -\frac{\tan A + \tan B}{1-\tan A\tan B}.$ Now take it from there ... .
• Jun 14th 2011, 12:18 PM
Zellator
Re: Two angles and the angle in between in 180 degrees; maximization
Quote:

Originally Posted by Opalg
$\tan(180^\circ-C) = -\tan C$, and so $\tan C = -\tan(A+B) = -\frac{\tan A + \tan B}{1-\tan A\tan B}.$ Now take it from there ... .

Hey Opalg!
I remember seeing this formula somewhere, but I didn't know how to put in motion.
So, it was easier than I thought. Thanks for your explanation!!!
All the best! :)