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

Hi Forum!

I have a question

https://lh6.googleusercontent.com/-u...Q6Y/angles.jpg

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.

Re: Two angles and the angle in between in 180 degrees; maximization

Quote:

Originally Posted by

**Zellator** Hi Forum!

I have a question

https://lh6.googleusercontent.com/-u...Q6Y/angles.jpg
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.

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

Re: Two angles and the angle in between in 180 degrees; maximization

Quote:

Originally Posted by

**Opalg** $\displaystyle \tan(180^\circ-C) = -\tan C$, and so $\displaystyle \tan C = -\tan(A+B) = -\frac{\tan A + \tan B}{1-\tan A\tan B}.$ 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! :)