Question Number 18167 by ajfour last updated on 16/Jul/17
Commented by ajfour last updated on 16/Jul/17
$$\mathrm{just}\:\mathrm{for}\:\mathrm{fun}. \\ $$
Commented by mrW1 last updated on 16/Jul/17
$$\emptyset=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{3}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{4}}=\mathrm{45}°−\mathrm{26}.\mathrm{6}°=\mathrm{18}.\mathrm{4}° \\ $$
Commented by ajfour last updated on 16/Jul/17
$$=\:\:\boldsymbol{\phi}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\mathrm{1}/\mathrm{2}}{\mathrm{1}+\mathrm{1}/\mathrm{2}}\right)=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\:. \\ $$