Question Number 61449 by Tawa1 last updated on 02/Jun/19
Commented by Tawa1 last updated on 02/Jun/19
$$\frac{\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} }\:\:=\:\:? \\ $$
Answered by perlman last updated on 02/Jun/19
$$\frac{{c}}{{sin}\left(\mathrm{30}\right)}=\frac{{a}}{{sin}\left(\mathrm{140}\right)}=\frac{{b}}{{sin}\left(\mathrm{130}\right)} \\ $$$${b}=\frac{{sin}\left(\mathrm{130}\right)}{{sin}\left(\mathrm{30}\right)}{c} \\ $$$${a}=\frac{{sin}\left(\mathrm{140}\right)}{{sin}\left(\mathrm{30}\right)}{c} \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }=\frac{{sin}^{\mathrm{2}} \left(\mathrm{130}\right)+{sin}^{\mathrm{2}} \left(\mathrm{140}\right)}{{sin}^{\mathrm{2}} \left(\mathrm{30}\right)}=\mathrm{4}\left({sin}^{\mathrm{2}} \left(\mathrm{130}\right)+{sin}^{\mathrm{2}} \left(\mathrm{140}\right)\right)=\mathrm{4}\left({sin}^{\mathrm{2}} \left(\mathrm{180}−\mathrm{50}\right)+{sin}^{\mathrm{2}} \left(\mathrm{90}+\mathrm{50}\right)\right) \\ $$$$=\mathrm{4}{sin}^{\mathrm{2}} \left(\mathrm{50}\right)+\mathrm{4}{cos}^{\mathrm{2}} \left(\mathrm{50}\right)=\mathrm{4}\left({sin}^{\mathrm{2}} \left(\mathrm{50}\right)+{cos}^{\mathrm{2}} \left(\mathrm{50}\right)\right)=\mathrm{4}×\mathrm{1}=\mathrm{4} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 02/Jun/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$