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 $\cos \angle BDC$?
+128474 See the following pic :
Let's find AC = √ (BC^2 - BA^2) = √ (15^2 -12^2) = √ [225 - 144] = √ 81 = 9
Because ABC is bisected, we have the folllwing relationship :
AD /DC = AB/ BC
AD/DC = 12/15 = 4/5
So
AC has 9 parts and AD is 4 of them....so AD = 4
And DC = 5
Using the Pythagorean Theorem
BD = √ (BA^2 + AD^2) = √ ( 12^2 + 4^2) = √ 160 = 4√ 10
Note that the sine of ADB = sine of BDC
sin ADB = AD / DB = 12 / (4√ 10) = 3/√ 10
And note that BDC is obtuse, so the cosine will be negative
And cos BDC = -√ [ 1 - ( sin ADB)^2] = √ [ 1 - (9/10)] = - √ ( 1/10) = -√ 10/10
CORRECTED !!!!
