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 \(\tan \angle BDC\)?

Guest Dec 12, 2018

#1**0 **

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

PartialMathematician Dec 13, 2018

#2**+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

CPhill Dec 13, 2018