Question Number 45187 by MrW3 last updated on 10/Oct/18 Commented by MrW3 last updated on 12/Oct/18 $${Solution}\:{to}\:{Q}\mathrm{45122}. \\ $$$${Find}\:{the}\:{length}\:{of}\:{path}\:{from}\:{A}\:{to}\:{B} \\ $$$${which}\:{has}\:{a}\:{constant}\:{slope}. \\ $$$$ \\ $$$${A}\:{point}\:{P}\:{on}\:{the}\:{path}\:{can}\:{be}\:{described}…
Question Number 176192 by adhigenz last updated on 14/Sep/22 $$\mathrm{Suppose}\:\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rectangle}.\:\mathrm{X}\:\mathrm{and}\:\mathrm{Y}\:\mathrm{are}\:\mathrm{points}\:\mathrm{on}\:\mathrm{BC}\:\mathrm{and}\:\mathrm{CD}\:\mathrm{respectively}, \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{ABX},\:\mathrm{CXY},\:\mathrm{and}\:\mathrm{AYD}\:\mathrm{are}\:\mathrm{3}\:\mathrm{cm}^{\mathrm{2}} ,\:\mathrm{4}\:\mathrm{cm}^{\mathrm{2}} ,\:\mathrm{and}\:\mathrm{5}\:\mathrm{cm}^{\mathrm{2}} \:\mathrm{respectively}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{AXY}. \\ $$ Answered by som(math1967) last updated on…
Question Number 176195 by adhigenz last updated on 14/Sep/22 Answered by behi834171 last updated on 15/Sep/22 $${BC}={BP}={PC}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}},{AB}={AR}={BR}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${PQ}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}−\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}×\mathrm{1}×{cos}\left(\mathrm{150}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}+\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow{PQ}=\sqrt{\frac{\mathrm{7}}{\mathrm{3}}} \\ $$$${cos}\measuredangle{CPQ}=\frac{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{7}}{\mathrm{3}}}\right)^{\mathrm{2}}…
Question Number 45122 by MrW3 last updated on 09/Oct/18 Commented by MrW3 last updated on 09/Oct/18 $${A}\:{mountain}\:{has}\:{the}\:{shape}\:{of}\:{a}\:{semi} \\ $$$${sphere}\:{with}\:{radius}\:{R}.\:{From}\:{a}\:{point}\:{A} \\ $$$${at}\:{the}\:{foot}\:{of}\:{mountain}\:{to}\:{a}\:{point}\:{B}\: \\ $$$${over}\:{point}\:{A}\:{two}\:{paths}\:{will}\:{be}\:{built}. \\ $$$${One}\:{path}\:{should}\:{have}\:{the}\:{shortest}…
Question Number 176100 by mnjuly1970 last updated on 12/Sep/22 $$ \\ $$$$\:\:\:\:\:{Q}\:: \\ $$$$\:\:\:\:\:{in}\:\:{A}\overset{\Delta} {{B}C}\::\:{A}\:=\:\mathrm{90}^{\mathrm{o}} \:\:\:{and}\:,\:{m}_{\:{b}} ^{\mathrm{2}} \:+\:{m}_{\:{c}} ^{\:\mathrm{2}} \:=\:\mathrm{100}.\:\:\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{find}\:\:{the}\:\:{value}\:{of}\:\:\:''\:\:\:{a}\:\:\:''\:. \\…
Question Number 176086 by amin96 last updated on 11/Sep/22 Commented by mahdipoor last updated on 11/Sep/22 $$?=\frac{\mathrm{1}}{\mathrm{2}{cosa}}\approx\mathrm{1}.\mathrm{4142} \\ $$$${a}={tan}^{−\mathrm{1}} \left(\frac{{sin}\frac{\mathrm{720}}{\mathrm{7}}+{sin}\frac{\mathrm{360}}{\mathrm{7}}−\mathrm{4}{sin}\left(\frac{\mathrm{90}}{\mathrm{7}}\right){sin}\left(\frac{\mathrm{900}}{\mathrm{7}}\right)}{{cos}\frac{\mathrm{360}}{\mathrm{7}}+{cos}\frac{\mathrm{720}}{\mathrm{7}}}\right) \\ $$$$ \\ $$ Terms…
Question Number 176076 by Agnibhoo98 last updated on 11/Sep/22 Commented by som(math1967) last updated on 11/Sep/22 $$\angle\boldsymbol{{B}}{AC}=\mathrm{90}° \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 44950 by ajfour last updated on 06/Oct/18 Commented by ajfour last updated on 06/Oct/18 Commented by ajfour last updated on 06/Oct/18 $${GF}\:=\:{CF}\:=\:{x} \\…
Question Number 176014 by cherokeesay last updated on 11/Sep/22 Commented by cherokeesay last updated on 11/Sep/22 $${The}\:{triangle}\:{ABC}\:{is}\:{right}-{angled}\:{at}\:{B}. \\ $$ Answered by cortano1 last updated on…
Question Number 44929 by kvbkvb last updated on 06/Oct/18 Commented by ajfour last updated on 06/Oct/18 $${See}\:{Q}.\mathrm{44395} \\ $$ Answered by rahul 19 last updated…