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Category: Coordinate Geometry

Question-25156

Question Number 25156 by tawa tawa last updated on 05/Dec/17 Answered by nnnavendu last updated on 05/Dec/17 $${we}\:{know}\:{that}\:{point}\:{on}\:{the}\:{x}\:−\:{asix}\:{A}\equiv\left(−{x},\mathrm{0}\right) \\ $$$${then}\:{from}\:{distance}\:{formula}\:{and}\:{diagram} \\ $$$${AB}={AD} \\ $$$$\sqrt{\left(−{x}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{0}−\mathrm{10}\right)^{\mathrm{2}}…

Question-155866

Question Number 155866 by cherokeesay last updated on 05/Oct/21 Answered by talminator2856791 last updated on 05/Oct/21 $$\:\mathrm{2}\sqrt{\mathrm{3}}\:+\:\mathrm{2}\sqrt{\mathrm{3}}\:−\:\frac{\mathrm{9}\pi\centerdot\mathrm{arctan}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{\mathrm{2}\pi}\:−\:\frac{\pi}{\mathrm{4}}\:−\:\frac{\mathrm{arctan}\left(\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)}{\mathrm{2}}\:\: \\ $$$$\:=\:\mathrm{4}\sqrt{\mathrm{3}}\:−\:\frac{\mathrm{9}}{\mathrm{2}}\:\mathrm{arctan}\left(\sqrt{\mathrm{3}}\right)−\:\frac{\pi}{\mathrm{4}}\:−\:\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\:=\:\mathrm{4}\sqrt{\mathrm{3}}\:−\:\frac{\mathrm{9}\pi}{\mathrm{6}}\:−\:\frac{\pi}{\mathrm{4}}\:−\:\frac{\pi}{\mathrm{6}} \\ $$$$\:=\:\mathrm{4}\sqrt{\mathrm{3}}\:−\:\frac{\mathrm{13}\pi}{\mathrm{12}} \\ $$…