In right triangle \(ABC\), we have \(\angle BAC = 90^\circ\) and \(D\) is on \(\overline{AC}\) such that \(\overline{BD}\) bisects \(\angle ABC\). If \(AB = 12\) and \(BC = 15\), then what is \(\cos \angle BDC\)?
+128475 B
12 15
A D C
We have a 9 - 12 - 15 right triangle
So AC = 9
Let AD = x and DC = 9 -x
And since BD bisects angle ABC,we have the following relationship
AB/ BC = AD / DC
12/ 15 = x / (9 - x) cross-multiply
12 ( 9 - x) = 15x
108 - 12x = 15x
108 = 27x
x = 4 = AD
And BD = √[ AD^2 + BA^2] = √[ 4^2 + 12^2 ] = √160 = 4√10
And since angles BDC and ADB are supplementary
cos (BDA) = -cos (BDC)
- cos(BDA) = cos(BDC)
And cos (BDA) = AD/ BD = 4 / [ 4√10] = 1 / √10 = √ 10 / 10
So cos (BDC) = - cos (BDA) = - √10/ 10
