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In the diagram, \(AB = 13\text{ cm},\) \( DC = 20\text{ cm},\) and \( AD = 5\text{ cm}.\) What is the length of  to the nearest tenth of a centimeter?


 Apr 21, 2018
 #1
avatar+972 
+1

You didn’t specify what we need to find, but assuming you want BC, here is my solution:

 

Using the Pythagorean Thereom, BD^2 = AB^2 - AD^2 = 13^2 - 5^2 = 12^2 

 

BD = 12

 

Using the same method, BC^2 = DC^2 - BD^2 = 20^2 - 12^2

 

You take the square root of that and you have your answer.

 Apr 21, 2018
 #2
avatar+98005 
+1

And if you want to find AC   ...

 

cos BDC  =  12/20  = 3/5

cos BDA  = 5/13

sin BDC = 4/5

sin BDA = 12/13

 

So  cos  (BDC +  BDA)  =  cosBDCcosBDA - sinBDC sinBDA =

                                            (3/5)(5/13) - (4/5)(12/13)  =

                                             15/65  -  48/65 =

                                              -33/65

 

So....using the  Law of Cosines

 

AC^2  = AD^2 + DC^2  - 2(AD * DC) cos ( BDC + BDA)

 

AC^2  = 5^2   +  20^2  - 2(5 * 20)(-33/65)

 

AC^2  =  425   + 200(33/65)

 

AC^2  = 6845 / 13

 

AC  ≈ 22.946

 

 

cool cool cool

 Apr 22, 2018

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