Angle Help!15 = arctan(2/x)  arctan (1/x)
1. The problem statement, all variables and given/known data
Basically, solve for x 15 = arctan(2/x)  arctan (1/x) 2. Relevant equations tan (AB) = (Tan A Tan B) / (1+Tan A*Tan B) 3. The attempt at a solution I really tried everything. My first step was to: Let y = arctan (2/x) Therefore, tan y = 2/x Similarly, u = 1/x Then, tan (yu) = (Tan y Tan u) / (1+Tan u*Tan y) = 15 15 = (2/x(1/x) / (1+(2/x^2) 15 = (1/x) / (x^2 + 2 / x^2) 15 = (x) / (x^2+2) 15x^2 + 30  x = 0 Which has no real roots :( But, with guess and check, it's around 0.65 
Re: Angle Help!15 = arctan(2/x)  arctan (1/x)
You need to post an attempt before we can help you. Start by taking the tangent of both sides.

Re: Angle Help!15 = arctan(2/x)  arctan (1/x)
Yeah, sorry. I just misclicked the first time

Re: Angle Help!15 = arctan(2/x)  arctan (1/x)
Alright, sorry guys to waste your time, but I believe I figured it out. Thanks for the hint of "tanning" both sides.
Instead of 15, it's supposed to be tan 15. So that, x/(x^2+2) = tan 15 x = 0.64 x = 3.08 (approximately) Thanks for the help! 
Re: Angle Help!15 = arctan(2/x)  arctan (1/x)
If you rearrange that equation you'll get the quadratic
[tex]x^2\frac{1}{tan(15)}x+2=0[/tex] There are no real solutions to this quadratic. 
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