Question Number 205403 by BaliramKumar last updated on 20/Mar/24 Answered by Rajpurohith last updated on 20/Mar/24 $${Clearly}\:\mathrm{cot}^{\mathrm{2}} \left({A}\right)\left(\mathrm{1}+\mathrm{cot}^{\mathrm{2}} \left({A}\right)\right)=\mathrm{1} \\ $$$${i}.{e},\mathrm{cot}^{\mathrm{2}} \left({A}\right)\left(\mathrm{cosec}^{\mathrm{2}} \left({A}\right)\right)=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{cos}^{\mathrm{2}}…
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Question Number 204946 by 0547 last updated on 02/Mar/24 $$\mathrm{2}+\mathrm{6}{i}×\mathrm{5}+\mathrm{4}{j} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 204845 by LuisTony last updated on 28/Feb/24 Answered by TonyCWX08 last updated on 29/Feb/24 $${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:…
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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…