The vertices of triangle\(ABC\) lie on the sides of equilateral triangle\(DEF\) , as shown. If \(CD=3\), \(CE=BD=2\), and \(\angle ACB =90^{\circ}\), then find \(AE\).
+14915 \(Find\ \overline{AE}.\)
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\(\overline{BC}=\sqrt{2^2+3^2-2\cdot 3\cdot cos\ 60°}\\ \overline{BC}=\sqrt{10}\)
\(\frac{sin\ BCD}{2}=\frac{sin\ 60°}{\sqrt{10}}\\ sin\ BCD=\frac{2\cdot sin\ 60°}{\sqrt{10}}\\ BCD=33.211°\)
\(ACE=180°-90°-33.211°\\ ACE=56.789°\)
\(EAC=180°-60°-56.789°\\ EAC=63.211°\)
\(\frac{\overline{AE}}{sin\ 56.789°}=\frac{2}{sin\ 63.211°}\\ \overline{AE}=\frac{2\cdot sin\ 56.789°}{sin\ 63.211°}\)
\(\overline{AE}=1.8745\)
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