Question Number 204845 by LuisTony last updated on 28/Feb/24 Answered by TonyCWX08 last updated on 29/Feb/24 $${Altura}\:{del}\:{arbol} \\ $$$$=\:\mathrm{95}\left(\mathrm{sin}\:\mathrm{25}\right) \\ $$$$\approx\:\mathrm{40}.\mathrm{15}{m} \\ $$$${Distancia}\:{de}\:{la}\:{cometa}\:{al}\:{suelo} \\ $$$$\approx\mathrm{40}.\mathrm{15}{m}…
Question Number 204729 by Amidip last updated on 26/Feb/24 Answered by A5T last updated on 26/Feb/24 $$\frac{{sin}\left(\mathrm{2}{x}\right)}{{AD}}=\frac{{sin}\left(\mathrm{140}−\mathrm{2}{x}\right)}{{AB}};\frac{{sin}\left(\mathrm{140}−\mathrm{3}{x}\right)}{{AD}}=\frac{{sin}\left({x}\right)}{{DC}={AB}} \\ $$$$\Rightarrow\frac{{sin}\left(\mathrm{140}−\mathrm{3}{x}\right)}{{sin}\left({x}\right)}=\frac{{sin}\left(\mathrm{2}{x}\right)}{{sin}\left(\mathrm{140}−\mathrm{2}{x}\right)}\Rightarrow{x}=\mathrm{35}° \\ $$ Terms of Service Privacy…
Question Number 204699 by AtulKumar last updated on 25/Feb/24 Commented by LuisTony last updated on 25/Feb/24 $${In}\:{spanish}? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 204654 by LuisTony last updated on 24/Feb/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 204663 by LuisTony last updated on 24/Feb/24 Answered by TonyCWX08 last updated on 25/Feb/24 $$\mathrm{sin}\left(\mathrm{52}°\mathrm{20}'\right)=\frac{{h}}{\mathrm{1200}} \\ $$$${h}=\mathrm{1200sin}\left(\mathrm{52}°\mathrm{20}'\right) \\ $$$${h}\approx\mathrm{949}.\mathrm{9}{m} \\ $$ Terms of…
Question Number 204628 by MM42 last updated on 23/Feb/24 $$ \\ $$ Answered by MM42 last updated on 23/Feb/24 Answered by A5T last updated on…
Question Number 204617 by Abdullahrussell last updated on 23/Feb/24 Commented by Frix last updated on 23/Feb/24 $$\mathrm{33} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 204471 by Abdullahrussell last updated on 18/Feb/24 Answered by TonyCWX08 last updated on 18/Feb/24 $$ \\ $$ Answered by MM42 last updated on…
Question Number 204293 by depressiveshrek last updated on 11/Feb/24 $$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{trig}\:\mathrm{identity}: \\ $$$$\frac{\mathrm{2sin}\alpha+\mathrm{sin3}\alpha+\mathrm{sin5}\alpha}{\mathrm{cos}\alpha−\mathrm{2cos2}\alpha+\mathrm{cos3}\alpha}=\frac{\mathrm{2cos2}\alpha}{\mathrm{tan}\frac{\alpha}{\mathrm{2}}} \\ $$ Commented by Frix last updated on 11/Feb/24 $$\mathrm{Try}. \\ $$$$\alpha=\frac{\pi}{\mathrm{6}}\:\Rightarrow\:\mathrm{lhs}=−\mathrm{5}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:\:\:\:\:\mathrm{rhs}=\mathrm{2}+\sqrt{\mathrm{3}} \\…
Question Number 204101 by necx122 last updated on 06/Feb/24 $${Find}\:{the}\:{possible}\:{values}\:{for}\:{x}\:{and}\:{y}\:{if} \\ $$$$\mathrm{10}{cosx}\:+\:\mathrm{12}{cos}\left({x}+{y}\right)=\mathrm{5} \\ $$$$\mathrm{10}{sinx}\:+\:\mathrm{12}{sin}\left({x}+{y}\right)=\mathrm{20}.\mathrm{66} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com