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Question Number 40615 by behi83417@gmail.com last updated on 24/Jul/18

Commented by behi83417@gmail.com last updated on 24/Jul/18

$$A\stackrel{}{CB}={90}^{\bullet },A\stackrel{}{CE}=E\stackrel{}{CM}=M\stackrel{}{CB}$$$$AC=CM=5,CB=12,MN\mathrm{\perp }AB.$$$$\mathit{M}\mathit{N}=?$$

Commented by ajfour last updated on 25/Jul/18

Answered by MJS last updated on 25/Jul/18

$$AB=\sqrt{5^2+{12}^2}=13$$$$\mathrm{\angle }ACM=60°$$$$AC=CM=5\Rightarrow AM=5$$$$\mathrm{\angle }CAB=\arcsin \frac{12}{13}$$$$\mathrm{\angle }MAB=\arcsin \frac{12}{13}-60°$$$$\mid MN\mid =5\sin (\arcsin \frac{12}{13}-60°)=\frac{30}{13}-\frac{25\sqrt{3}}{26}\approx .642259$$

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