Question Number 196972 by otchereabdullai@gmail.com last updated on 05/Sep/23
Answered by Frix last updated on 05/Sep/23
$${t}=\mathrm{tan}\:\theta \\ $$$$\frac{\mathrm{2}}{{t}^{\mathrm{2}} +\mathrm{1}}\left(\mathrm{14}{t}^{\mathrm{2}} +\mathrm{9}{t}−\mathrm{14}\right)=\mathrm{0} \\ $$$${t}=−\frac{\mathrm{9}}{\mathrm{28}}\pm\frac{\sqrt{\mathrm{865}}}{\mathrm{28}} \\ $$$$\theta=\mathrm{tan}^{−\mathrm{1}} \:\frac{−\mathrm{9}+\sqrt{\mathrm{865}}}{\mathrm{28}}\:\approx\mathrm{36}.\mathrm{091}° \\ $$
Commented by otchereabdullai@gmail.com last updated on 05/Sep/23
$${thanks} \\ $$