Question Number 193548 by lmcp1203 last updated on 16/Jun/23
Commented by lmcp1203 last updated on 16/Jun/23
$${hint}\:{by}\:{trigonometry}\:\:{x}=\mathrm{10} \\ $$
Answered by Subhi last updated on 16/Jun/23
$$ \\ $$$$\frac{{a}}{{sin}\left(\mathrm{30}\right)}=\frac{{b}}{{sin}\left(\mathrm{100}\right)}\:\Rrightarrow\:{a}=\frac{{b}}{\mathrm{2}{sin}\left(\mathrm{100}\right)}\:\left({i}\right) \\ $$$$\frac{{b}}{{sin}\left({x}\right)}=\frac{{c}}{{sin}\left(\mathrm{10}\right)}\:\:\Rrightarrow\:{c}=\frac{{sin}\left(\mathrm{10}\right){b}}{{sin}\left({x}\right)}\:\left({ii}\right) \\ $$$$\frac{{a}}{{sin}\left(\mathrm{20}−{x}\right)}=\frac{{c}}{{sin}\left(\mathrm{20}\right)}\:\Rrightarrow\:\frac{{a}}{{c}}=\frac{{sin}\left(\mathrm{20}−{x}\right)}{{sin}\left(\mathrm{20}\right)}=\frac{{i}}{{ii}} \\ $$$$\frac{{sin}\left({x}\right)}{\mathrm{2}{sin}\left(\mathrm{100}\right).{sin}\left(\mathrm{10}\right)}=\frac{{sin}\left(\mathrm{20}−{x}\right)}{{sin}\left(\mathrm{20}\right)}=\frac{{sin}\left(\mathrm{20}\right){cos}\left({x}\right)−{sin}\left({x}\right){cos}\left(\mathrm{20}\right)}{{sin}\left(\mathrm{20}\right)} \\ $$$$\mathrm{2}{sin}\left(\mathrm{20}\right){sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right){cos}\left({x}\right)−\mathrm{2}{sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right){cos}\left(\mathrm{20}\right){sin}\left({x}\right)={sin}\left(\mathrm{20}\right){sin}\left({x}\right) \\ $$$$\left({sin}\left(\mathrm{20}\right)+\left(\mathrm{2}{sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right){cos}\left(\mathrm{20}\right)\right){sin}\left({x}\right)=\left(\mathrm{2}{sin}\left(\mathrm{20}\right){sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right)\right){cos}\left({x}\right)\right. \\ $$$${x}={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{sin}\left(\mathrm{20}\right){sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right)}{{sin}\left(\mathrm{20}\right)+\left(\mathrm{2}{sin}\left(\mathrm{100}\right){sin}\left(\mathrm{10}\right){cos}\left(\mathrm{20}\right)\right)}\right)=\mathrm{10} \\ $$
Commented by Subhi last updated on 16/Jun/23
Commented by lmcp1203 last updated on 16/Jun/23
$${thanks}. \\ $$