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In ΔBCD, \overline{BD} BD is extended through point D to point E, m∠CDE = (6x-13)^{\circ}(6x−13) ∘ , m∠BCD = (2x+10)^{\circ}(2x+10) ∘ , and m∠DBC = (x+1)^{\circ}(x+1) Find m∠CDE.

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sqdancefan

Answer:

  m∠CDE = 35°

Step-by-step explanation:

The exterior angle CDE is equal to the sum of the opposite interior angles, BCD and DBC. So, you have ...

  6x -13 = (2x+10) +(x +1)

  3x -24 = 0 . . . . . . . . . . . . subtract the right side, simplify

  x -8 = 0 . . . . . . . . . . . . . . divide by 3

  x = 8 . . . . . . . . . . . . . . . . . add 8

  6x -13 = 6·8 -13 = 35 . . . . substitute for x in the expression for CDE

The measure of angle CDE is 35°.

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