In triangle ABC with AB = 10, let D be a point on side BC such that AD bisects angle BAC. If CD/BD = 2 and the area of ABC is 50, compute the value of BAD in degrees
+130555 Since AD bisects angle BAC....we have this relationship
AC / CD = AB / BD
AC / [ 2BD ] = AB / BD
AC / 2 = AB
AC / 2 = 10
AC = 20
And we have this :
area of ABC = (1/2)(AB)(AC) sin BAC ....so....
50 = (!/2) (10)(20) sin BAC
100 = 200 sin BAC divide both sides by 200
1/2 = sin BAC
Using the arcsin we have that
arcsin (BAC) = 1/2 = 30°
And since AD bisects BAC.....then BAD = 15°
