+312 In right triangle ABC, we have angle BAC = 90 degrees and D is on AC such that BD bisects ABC. If AB = 12 and BC = 15, then what is the cosine of angle BDC?
Draw the triangle ABC, Plot D on AC and draw a line from angle ABC to point D, Thus, ABD=DBC
In the right angle triangle ABC, we know that AB=12, BC=15 therefore, AC=9
We have all sides so we can use either sine or cosine or tangent to find angle ABC
let's use cosine
Let, angle ABC=y
cos(y)=12/15
y=cos^-1(12/15) = 36.869897645844
Since BD bisects ABC
Therefore BDC=BDA as we said earlier and we know that DBC+DBA=ABC thus, BDC= 1/2ABC
so BDC=(1/2)36.869897645844 = 18.434948822922
We know 2 angles in a triangle so the third angle
180-(36.869897645844+90)=53.130102354156 (Angle BCA)
In triangle BDC, we know 2 angles which are Angle DBC=18.434948822922 and angle BCA=53.1301023541456
Therefore angle BDC=180-(53.1301023541456+18.434948822922) = 108.4349488229324
cos(108.4349488229324) = -0.316227766017
Note finding angle BDC is easier if we just added angle BAD+DBA since BDC is exterior angle of triangle ABD.
+129852 See the image
AB = 12 BC = 15 so AC = 9
Since BD bisects angle ABC, then
AD / DC = AB/ BC
AD / DC = 12/15
AD / DC = 4/5
So AD = (4/9)(9) = 4
So D =(4,0) and DC = AC - AD = 9 - 4 = 5
And the distance from B to D = sqrt (12^2 + 4^2 ) = sqrt (160)
So....using the Law of Cosines....
BC^2 = DC^2 + BD^2 -2(DC* BD)cos(BDC)
15^2 = 5^2 + 160 - 2 ( 5sqrt(160)) cos(BDC)
40 = -10sqrt (160) cos (BDC)
-40 / (10 sqrt (160) ) = cos (BDC)
-4 / [ sqrt (160) ] =cos (BDC)
-4sqrt (160) / 160 = cos (BDC)
-sqrt (160) / 40 = cos (BDC)
- sqrt (10) / 10 = cos (BDC)
