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March 3, 2015

March 3, 2015

Posted by **Pooja** on Wednesday, August 10, 2011 at 5:30am.

- geometry -
**Reiny**, Wednesday, August 10, 2011 at 8:17amSince AD is the median

BD = DC = 9

In ∆ABD, let ∢BDA = Ø

by the cosine law:

8^2 = 9^2 + 7^2 - 2(9)(7)cosØ

cosØ = .5238

Ø = 58.4°

the ∢ADC = 121.59°

now in ∆ADC

AC^2 = 49 + 64 - 2(7)(8)cos121.59

= 171.666667

AC = 13.102 or 13

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