+1207 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 \(\triangle ABC\)?
+129847 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
