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In triangle ABC, AB = BC, AC = 16, and angle B = 90 degrees.  AC is extended to D such that BD = 17.  Find CD.

 

 May 14, 2020
 #1
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Note  that  if AB  = BC......then angles CAB  and BAC  = 45°

 

Then  let AB, BC  = x

 

And we have  that

 

x^2  + x^2  = 16^2

 

2x^2  = 256

 

x^2   = 128

 

x = 8√2

 

Then   each of these  sides  =   8/√2

 

Looking at triangle BCD,  angle DCB  = 180 - 45  =   135°

 

Using the Law of Sines , we have that

 

sin CDB  /(8/√2 )  = sin DCB /  17

 

sin CDB  /  (8/√2)  =  sin 135 / 17

 

sin CDB  = (8/√2) (1/√2) / 17

 

sin CDB =  8/17

 

Then cos  CDB  =  15/17

 

Using  the Law  of Cosines

 

AB^2   = BD^2  +  (16 + CD)^2 - 2(BD ( 16 + CD))cos CDB

 

(8√2)^2  =  17^2  + 256 + 32CD + CD^2  - 2 (17(16 + CD))(15/17)

 

128 = 289 + 256 + 32CD +CD^2  - 2 (15) (16 + CD)

 

-417  =  32CD + CD^2  - 480 - 30CD

 

63  = CD^2 + 2CD

 

CD^2 + 2CD - 63  = 0   factor

 

(CD + 9) (CD - 7)  = 0

 

The second factor  gives  us the answer

 

CD - 7  = 0

 

CD  = 7

 

 

cool cool cool

 May 14, 2020

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