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Category: Geometry

Question-193906

Question Number 193906 by cherokeesay last updated on 22/Jun/23 Answered by Subhi last updated on 22/Jun/23 $$ \\ $$$$\frac{\mathrm{1}}{{sin}\left(\mathrm{18}\right)}=\frac{\mathrm{1}+{y}}{{sin}\left(\mathrm{144}−{x}\right)}\: \\ $$$$\frac{\mathrm{1}}{{sin}\left(\mathrm{18}\right)}=\frac{{y}}{{sin}\left({x}\right)}\:\Rrightarrow\:{y}\:=\:\frac{{sin}\left({x}\right)}{{sin}\left(\mathrm{18}\right)} \\ $$$$\frac{\mathrm{1}}{{sin}\left(\mathrm{18}\right)}=\frac{\mathrm{1}+\frac{{sin}\left({x}\right)}{{sin}\left(\mathrm{18}\right)}}{{sin}\left(\mathrm{144}−{x}\right)} \\ $$$${sin}\left(\mathrm{144}−{x}\right)={sin}\left(\mathrm{18}\right)+{sin}\left({x}\right)…

Question-193726

Question Number 193726 by mr W last updated on 18/Jun/23 Answered by som(math1967) last updated on 18/Jun/23 $${a}+{b}+{c} \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\:+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \mathrm{1} \\ $$$$=\mathrm{tan}^{−\mathrm{1}}…

Question-193635

Question Number 193635 by mr W last updated on 17/Jun/23 Commented by mr W last updated on 17/Jun/23 $${find}\:{the}\:{largest}\:{circle}\:{and}\:{the}\:{largest} \\ $$$${square}\:{which}\:{you}\:{can}\:{completely} \\ $$$${cover}\:{with}\:{three}\:{circular}\:{plates}\:{with} \\ $$$${radius}\:\mathrm{1}\:{respectively}.…

Question-193522

Question Number 193522 by Mingma last updated on 16/Jun/23 Answered by Subhi last updated on 16/Jun/23 $$\frac{{x}+{z}}{{sin}\left(\mathrm{100}\right)}=\frac{{x}}{{sin}\left(\mathrm{40}\right)} \\ $$$${sin}\left(\mathrm{100}\right){x}={sin}\left(\mathrm{40}\right){x}+{sin}\left(\mathrm{40}\right){z}\:\Rrightarrow\:{x}=\frac{{sin}\left(\mathrm{40}\right){z}}{{sin}\left(\mathrm{10}\right)−{sin}\left(\mathrm{40}\right)}\:\left({i}\right) \\ $$$$\frac{{z}}{{sin}\left({y}−\mathrm{40}\right)}=\frac{{x}+{z}}{{sin}\left({u}\right)} \\ $$$${u}+{y}−\mathrm{40}=\mathrm{180}−\mathrm{140}=\mathrm{40}\:\Rrightarrow\:{y}=\mathrm{80}−{u}\:\left({ii}\right) \\ $$$$\frac{{z}}{{sin}\left(\mathrm{40}−{u}\right)}=\frac{{x}+{z}}{{sin}\left({u}\right)}…