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In right triangle $ABC$, we have $\angle BAC = 90^\circ$ and $D$ is the midpoint of $\overline{AC}$. If $AB = 7$ and $BC = 7 \sqrt{3}$, then what is $\tan \angle BDC$?

 Sep 8, 2023
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B

 

7                 7sqrt (3)  =  sqrt (147)

 

A           D            C

 

AC =  sqrt  [ 147 - 49 ]  = sqrt [98 ] =  7sqrt (2)

DC = 3.5sqrt (2) = AD  =  sqrt [24.5]

 

BD  =  sqrt [ 7^2 + 24.5]  = sqrt [ 73.5] 

 

Using Law of Cosines

 

BC^2  = BD^2  + DC^2  -  2(BD* DC) cos (BDC)

 

147  = 73.5 + 24.5 - 2 [ sqrt [ 73.5] * sqrt [ 24.5] ] cos (BDC)

 

49 / [- 2 sqrt (7203 / 4 )  ] =    cos (BDC)

 

-49 / sqrt(7203)  = cos (BDC)

 

sin(BDC) =   sqrt [  1 - (49)^2 / 7203 ]  =  sqrt (6) /  3

 

tan BDC =  [sqrt (6) / 3 ]  / [ -49 / sqrt (7203) ] =    [sqrt(6)*sqrt (7203) / [ -49 * 3]  =  

 

(sqrt 2) (sqrt (3) * 49 * sqrt  (3) )  / [ -147]  = 

 

-sqrt (2) * 147  /  147  = 

 

   -sqrt (2) 

 

cool cool cool

 Sep 9, 2023

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