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In \(\triangle ABC\), we have \(\angle BAC = 60^\circ\) and \(\angle ABC = 45^\circ\). The bisector of \(\angle A\) intersects \(\overline{BC}\) at point \(T\), and \(AT = 24\). What is the area of \(\triangle ABC\)?


 

 Mar 2, 2019
 #1
avatar+104647 
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Angle TAB =  105°

 

Using the Law of Sines

 

AB / sin ATB = AT/ sin CBA

 

AB / sin 105 = AT / sin 45

 

AB =  24 * sin 105 /sin 45 =  24 (1/2) ( 1 + sqrt(3) ) =  12 (1 + sqrt (3)  )

 

And using it again

 

AC / sin CBA = AB / sin ACB

 

AC / (sqrt (2) /2)  = 12 ( 1 + sqrt (3) )  

 

AC =  6sqrt (2) (1 + sqrt 3 )

 

 

So...the area of  ABC  =

 

(1/2) (AB) (AC) * sin (60) =

 

(1/2) 12 ( 1 + sqrt (3) ) * 6sqrt (2) (1 + sqrt (3) )  * sqrt (3) /2  =

 

18 sqrt (6) * ( 1 + sqrt 3)^2  =

 

18 sqrt (6)  ( 1 + 2sqrt(3) + 3) =

 

18sqrt (6) (4 + 2sqrt(3) ) =

 

72sqrt (6) + 108sqrt(2)  units ^2  =  

 

36sqrt(2) ( 2sqrt(3) + 3)   units^2

 

 

cool cool cool

 Mar 2, 2019
edited by CPhill  Mar 2, 2019

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