Any chance of solving for $\displaystyle \theta_2$ in either cartesian or polar form? $\displaystyle \ \cos\theta_1=\sqrt{\frac{(\pi+2\theta_2(b-1))^2}{\pi^2 (1+\tan^2{\theta_2})}}$ Thanks
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Originally Posted by rainer Any chance of solving for $\displaystyle \theta_2$ in either cartesian or polar form? $\displaystyle \ \cos\theta_1=\sqrt{\frac{(\pi+2\theta_2(b-1))^2}{\pi^2 (1+\tan^2{\theta_2})}}$ Thanks Hi $\displaystyle 1+\tan^2{\theta_2} = \frac{1}{\cos^2{\theta_2}}$ $\displaystyle \pi\ \cos\theta_1=|(\pi+2\theta_2(b-1))\cos{\theta_2}|$ Unless you have additional information on $\displaystyle \theta_2$ you cannot solve without calculator
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