Question Number 143586 by mim24 last updated on 16/Jun/21
Commented by muallim_riyoziyot last updated on 16/Jun/21
$${xaa} \\ $$
Answered by bobhans last updated on 16/Jun/21
$$\mathrm{tan}\:\theta\:=\:\mid\frac{−\frac{\mathrm{2}}{\mathrm{3}}−\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{1}+\left(−\frac{\mathrm{2}}{\mathrm{3}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\mid\frac{\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{1}+\mathrm{1}}\mid=\:\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\:\Rightarrow\theta\:=\:\mathrm{arctan}\:\left(\frac{\mathrm{5}}{\mathrm{12}}\right)=\mathrm{0}.\mathrm{394791} \\ $$