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What are the lengths AE and CE?

 

 Jul 21, 2022
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Because Triangle ADB is a right triangle, DB = \(\sqrt{81-16} = \sqrt{65}\)

 

\(sin^{-1}(\frac{1}{2}) = \angle{DCA} = 30\)\(DC = 4\sqrt{3}\).

 

\(\angle{B} \cong \angle{B}, \angle{D} = \angle{E}, \) Triangle BDA is congruent to Triangle BEC. 

 

\(\frac{x}{CB} = \frac{4}{9} = sin(\angle{B})\),

 

 \(CB = \frac{16\sqrt{3}+4\sqrt{65}}{9}\). Use the Pythagorean Theorem to solve for AE. 

 

Tried my best. Please correct me if I did anything wrong. smiley

 Jul 21, 2022
edited by nao173  Jul 21, 2022

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