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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=25, then what is \(\tan \angle BDC\)?

 Dec 12, 2018
 #1
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Is there a diagram? (my bad, i didnt know that a diagram was not needed, aka my trig skills = not that great)

 Dec 13, 2018
edited by PartialMathematician  Dec 13, 2018
 #3
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I agree, PM....a diagram IS often helpful...!!!!!

 

 

 

cool cool cool

CPhill  Dec 13, 2018
 #2
avatar+101871 
+1

B

                  

7                25

 

A          D    12      C

 

AC =  sqrt ( 25^2 - 7^2]  =  sqrt ( 625 - 49)   =  sqrt (576) = 24

 

And since D is the midpoint of AC, then DC =  (1/2)24 = 12  = AD

 

And tan BDA =  7/12

 

And angle  BDC is supplemental to  angle BDA

 

So.....BDC =  180° - BDA

 

So.....letting BDA = theta  and using  a trig identity  we have

 

tan ( 180° - theta)  =  tan (BDC) - tan (theta) = - tan(BDA) =  -7/12

 

 

 

cool cool cool

 Dec 13, 2018

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