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# Trigonometry

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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.

Jul 21, 2022
edited by nao173  Jul 21, 2022