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?
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