Question Number 41459 by ajfour last updated on 07/Aug/18 Commented by ajfour last updated on 07/Aug/18 $${how}\:{to}\:{draw}\:{without}\:{solving}\:? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 106990 by I want to learn more last updated on 08/Aug/20 $$\mathrm{Two}\:\mathrm{places}\:\:\mathrm{A}\:\:\mathrm{and}\:\:\mathrm{B}\:\:\mathrm{both}\:\mathrm{on}\:\mathrm{a}\:\mathrm{parallel}\:\mathrm{of}\:\mathrm{latitude}\:\:\alpha°\mathrm{N} \\ $$$$\mathrm{differs}\:\mathrm{in}\:\mathrm{longitudes}\:\mathrm{by}\:\:\theta°.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{distance} \\ $$$$\mathrm{between}\:\mathrm{them}\:\mathrm{is}:\:\:\:\:\:\frac{\left[\mathrm{2}\:\mathrm{sin}^{−\:\mathrm{1}} \left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right)\right]}{\mathrm{360}}\:\:×\:\:\mathrm{2}\pi\mathrm{R} \\ $$$$ \\ $$$$\mathrm{Topic}:\:\:\mathrm{Longitude}\:\mathrm{and}\:\mathrm{Latitude} \\ $$…
Question Number 106977 by mr W last updated on 08/Aug/20 Answered by mr W last updated on 08/Aug/20 $${A}\left({a},\mathrm{0}\right)\:{and}\:{B}\left({b},\mathrm{0}\right) \\ $$$${P}\:{lies}\:{on}\:{y}={x}^{\mathrm{2}} \\ $$$${find}\:{the}\:{minimum}\:{of}\:{perimeter}\:{of} \\ $$$$\Delta{PAB}.…
Question Number 172504 by mnjuly1970 last updated on 28/Jun/22 Answered by mr W last updated on 28/Jun/22 Commented by mr W last updated on 28/Jun/22…
Question Number 172484 by mnjuly1970 last updated on 27/Jun/22 Answered by mr W last updated on 28/Jun/22 $${let}\:\theta=\angle{ADC} \\ $$$$\frac{\mathrm{10}}{\mathrm{sin}\:\theta}=\frac{\mathrm{5}}{\mathrm{sin}\:\alpha}=\frac{{AD}}{\mathrm{sin}\:\left(\theta+\alpha\right)} \\ $$$$\frac{\mathrm{6}}{\mathrm{sin}\:\mathrm{2}\alpha}=\frac{{AD}}{\mathrm{sin}\:\left(\theta−\mathrm{2}\alpha\right)} \\ $$$$\mathrm{sin}\:\theta=\mathrm{2}\:\mathrm{sin}\:\alpha \\…
Question Number 172483 by mnjuly1970 last updated on 27/Jun/22 Answered by mr W last updated on 28/Jun/22 Commented by mr W last updated on 28/Jun/22…
Question Number 172453 by mr W last updated on 27/Jun/22 Commented by infinityaction last updated on 27/Jun/22 $$\mathrm{22}\:???? \\ $$ Commented by mr W last…
Question Number 172437 by infinityaction last updated on 26/Jun/22 Answered by mr W last updated on 27/Jun/22 $${R}={radius}\:{of}\:{semicircle} \\ $$$${r}={radius}\:{of}\:{circle} \\ $$$${O}={center}\:{of}\:{semicircle} \\ $$$${AC}=\mathrm{2}{R}\:\mathrm{cos}\:\theta \\…
Question Number 41348 by behi83417@gmail.com last updated on 06/Aug/18 Commented by MJS last updated on 06/Aug/18 $$\mathrm{trapezoid}\:{ABCD} \\ $$$${a}={AB}\:\:{b}={BC}\:\:{c}={CD}\:\:{d}={DA} \\ $$$${e}={AC}\:\:{f}={BD} \\ $$$$\mathrm{which}\:\mathrm{triangles}\:\mathrm{do}\:\mathrm{you}\:\mathrm{mean}? \\ $$$$\bigtriangleup{ABC}\:\mathrm{and}\:\bigtriangleup{ABD}?…
Question Number 41347 by behi83417@gmail.com last updated on 06/Aug/18 Answered by MJS last updated on 06/Aug/18 $${m}_{{a}} =\sqrt{\frac{{b}^{\mathrm{2}} }{\mathrm{2}}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${m}_{{b}} =\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+\frac{{c}^{\mathrm{2}}…