# Math Help - Help finding x

1. ## Help finding x

Hi can anybody show me how to find x?

-180= -90 - tan^-1 (x/16) - tan^-1 (x/200)

2. ## There are multiple ways to do this

Originally Posted by al2308
Hi can anybody show me how to find x?

-180= -90 - tan^-1 (x/16) - tan^-1 (x/200)
There is no alebraic way I am away of but you could do this... Say $f(x)=90-arctan\bigg(\frac{x}{16}\bigg)-arctan\bigg(\frac{x}{200}\bigg)$ then we want $f(x)=0$...I would graph...or use the intermediate value theorem...or just use a graphing calculator

3. Hello,

This is straaaaaaaaaaaaaange

90=arctan(x/16)+arctan(x/200)

If we compose with tan, we have :

tan(90)=tan(arctan(x/16)+arctan(x/200))

But tan(90)= infinity :'(

We can find somewhere the general formula for tan(a+b). I haven't looked further, maybe there is a solution though.

4. Originally Posted by Moo
Hello,

This is straaaaaaaaaaaaaange

90=arctan(x/16)+arctan(x/200)

If we compose with tan, we have :

tan(90)=tan(arctan(x/16)+arctan(x/200))

But tan(90)= infinity :'(

We can find somewhere the general formula for tan(a+b). I haven't looked further, maybe there is a solution though.
He is right I am so used to putting cookie cutter responses to find the x...since $tan\bigg(\frac{\pi}{2}\bigg)=\infty$ and the image of $arctan(x)$ is $\bigg(\frac{-\pi}{2},\frac{\pi}{2}\bigg)$...therefore there is no solutions..that is unless this is in radians and not degrees? wait once again...I didnt look at this...I didnt read it right...sory

5. Originally Posted by al2308
Hi can anybody show me how to find x?

-180= -90 - tan^-1 (x/16) - tan^-1 (x/200)
You can solve this numerically (assuming that 180 and 90 are in degrees) to get x ~= 56.57

RonL

6. Originally Posted by Moo
Hello,

This is straaaaaaaaaaaaaange

90=arctan(x/16)+arctan(x/200)

If we compose with tan, we have :

tan(90)=tan(arctan(x/16)+arctan(x/200))

But tan(90)= infinity :'(

We can find somewhere the general formula for tan(a+b). I haven't looked further, maybe there is a solution though.
Not significant because (slipping into radians) for instance:

$\pi/2=\arctan(\tan(\pi/4))+\arctan(\tan(\pi/4))$

and yet:

$\tan(\pi/2)=\tan(\arctan(\tan(\pi/4))+\arctan(\tan(\pi/4)))=\infty$

RonL

7. Hm so there can be an analytic solution, or i really misunderstood your message

Edit : ok, got it

8. Need verification (can't sleep until i find a correct method ), pleaaase, what d'you think of that ?

(little mistake, it's 40 sqrt(2))

(hey, it's the same as galactus's answer !)

9. Originally Posted by Moo
Hm so there can be an analytic solution, or i really misunderstood your message

Edit : ok, got it

Well there could be a closed form solution in terms of elementary functions, but it is unlikely unless the problem has some special structure, but I can't see any here.

(someone will be along in a minute to prove me wrong in this case - it always happens).

RonL

10. Originally Posted by Moo
Need verification (can't sleep until i find a correct method ), pleaaase, what d'you think of that ?

(little mistake, it's 40 sqrt(2))

(hey, it's the same as galactus's answer !)
Hey, you changed the answer before I could correct it.

RonL

11. So is it ok this way ?

PS : if you can reduce the size of the image, it would be wonderful as i have no idea how to do ^^'