+0  
 
0
64
2
avatar

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

 Mar 5, 2020
 #1
avatar+109499 
+1

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°

 

 

cool cool cool

 Mar 5, 2020
 #2
avatar
0

You did a good job here !!! smiley

Guest Mar 6, 2020

53 Online Users

avatar
avatar
avatar
avatar
avatar
avatar