+0  
 
0
50
2
avatar

Angle BAC= 60 and agnle ABC = 45. Angle bisector of BAC meets BC at T.  If AT = 40, what is the area of triangle ABC?

 

 Dec 7, 2020
 #1
avatar+114221 
+2

Angle CAT  = 30   Angle TAB  = 30  Angle ABC  = 45   Angle ACB  = 180 - 60 - 45  = 75

So angle  ATB   = 180  - 30 - 45 =   105

So angle  ATC  =180 - 105  =  75

 

Using the  Law of  Sines we have  that

 

AT / sin ABC =   AB /sin ATB

 

40 / sin 45  = AB/ sin 105

 

AB  =  40 sin105 / sin45  = 20 ( 1 + sqrt (3) )

 

Since angle  ACB  = angle ATC.....then  AT =  AC = 40

 

So area  of ABC  =  (1/2) AC * AB  sin (CAB)   =  

 

(1/2) 40 [ 20 (1 +  sqrt (3) ) ] [ sin 60]  =

 

20 [ 20 (1 + sqrt (3)  ] sqrt (3)  / 2  =

 

200 ( 1 + sqrt (3) )  (sqrt (3) )  =

 

200 ( 3 + sqrt (3) )  ≈  946.4 units^2

 

cool cool cool

 Dec 7, 2020
 #2
avatar+846 
+2

Angle BAC= 60 and angle ABC = 45. The angle bisector of BAC meets BC at T.  If AT = 40, what is the area of triangle ABC?

 

CF = TF = sin(15º) * 40 

 

AF = cos(15º) * 40

 

[ABC] = 1/2 (AF*CF + BF*AF)

 

 

 Dec 7, 2020

29 Online Users

avatar
avatar
avatar
avatar