Question Number 27254 by tawa tawa last updated on 03/Jan/18
Answered by mrW1 last updated on 04/Jan/18
![(a) α=28.7−21.3=7.4°=((7.4π)/(180)) [rad] β=38.6° R=6400 km L_(PQ) =αRcos β=((7.4π×6400×cos 38.6°)/(180))=646 km (b) Δβ=38.6−28.4=10.2°=((10.2π)/(180)) [rad] L_(PT) =ΔβR=((10.2π×6400)/(180))=1140 km (c) t=(L/v)=((646+1140)/(650))=2.75h=2h45m](https://www.tinkutara.com/question/Q27268.png)
$$\left({a}\right) \\ $$$$\alpha=\mathrm{28}.\mathrm{7}−\mathrm{21}.\mathrm{3}=\mathrm{7}.\mathrm{4}°=\frac{\mathrm{7}.\mathrm{4}\pi}{\mathrm{180}}\:\left[{rad}\right] \\ $$$$\beta=\mathrm{38}.\mathrm{6}° \\ $$$${R}=\mathrm{6400}\:{km} \\ $$$${L}_{{PQ}} =\alpha{R}\mathrm{cos}\:\beta=\frac{\mathrm{7}.\mathrm{4}\pi×\mathrm{6400}×\mathrm{cos}\:\mathrm{38}.\mathrm{6}°}{\mathrm{180}}=\mathrm{646}\:{km} \\ $$$$ \\ $$$$\left({b}\right) \\ $$$$\Delta\beta=\mathrm{38}.\mathrm{6}−\mathrm{28}.\mathrm{4}=\mathrm{10}.\mathrm{2}°=\frac{\mathrm{10}.\mathrm{2}\pi}{\mathrm{180}}\:\left[{rad}\right] \\ $$$${L}_{{PT}} =\Delta\beta{R}=\frac{\mathrm{10}.\mathrm{2}\pi×\mathrm{6400}}{\mathrm{180}}=\mathrm{1140}\:{km} \\ $$$$ \\ $$$$\left({c}\right) \\ $$$${t}=\frac{{L}}{{v}}=\frac{\mathrm{646}+\mathrm{1140}}{\mathrm{650}}=\mathrm{2}.\mathrm{75}{h}=\mathrm{2}{h}\mathrm{45}{m} \\ $$
Commented by tawa tawa last updated on 04/Jan/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$