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Question-40615




Question Number 40615 by behi83417@gmail.com last updated on 24/Jul/18
Commented by behi83417@gmail.com last updated on 24/Jul/18
AC^� B=90^• ,AC^� E=EC^� M=MC^� B  AC=CM=5,CB=12,MN⊥AB.  MN=?
$${A}\overset{} {{C}B}=\mathrm{90}^{\bullet} ,{A}\overset{} {{C}E}={E}\overset{} {{C}M}={M}\overset{} {{C}B} \\ $$$${AC}={CM}=\mathrm{5},{CB}=\mathrm{12},{MN}\bot{AB}. \\ $$$$\boldsymbol{{MN}}=? \\ $$
Commented by ajfour last updated on 25/Jul/18
Answered by MJS last updated on 25/Jul/18
AB=(√(5^2 +12^2 ))=13  ∠ACM=60°  AC=CM=5 ⇒ AM=5  ∠CAB=arcsin ((12)/(13))  ∠MAB=arcsin ((12)/(13)) −60°  ∣MN∣=5sin(arcsin ((12)/(13)) −60°)=((30)/(13))−((25(√3))/(26))≈.642259
$${AB}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\mathrm{13} \\ $$$$\angle{ACM}=\mathrm{60}° \\ $$$${AC}={CM}=\mathrm{5}\:\Rightarrow\:{AM}=\mathrm{5} \\ $$$$\angle{CAB}=\mathrm{arcsin}\:\frac{\mathrm{12}}{\mathrm{13}} \\ $$$$\angle{MAB}=\mathrm{arcsin}\:\frac{\mathrm{12}}{\mathrm{13}}\:−\mathrm{60}° \\ $$$$\mid{MN}\mid=\mathrm{5sin}\left(\mathrm{arcsin}\:\frac{\mathrm{12}}{\mathrm{13}}\:−\mathrm{60}°\right)=\frac{\mathrm{30}}{\mathrm{13}}−\frac{\mathrm{25}\sqrt{\mathrm{3}}}{\mathrm{26}}\approx.\mathrm{642259} \\ $$

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