In Triangle ABC, we have angle BAC is 60 degrees and angle ABC is 45 degrees. The bisector of angle A intersects BC at point T , and AT=24. What is the area of Triangle ABC?
+130511
If AT is an angle bisector of BAC, then BAT = 30° and ATB = 180 - 30 - 45 = 105°
We can find AB as follows
AT/sin45 = AB/sinATB
24/sin45 = AB/sin105
AB = 24sin105/sin45
And angle ATC is supplementary to angle ATB = 180 - 105 = 75°
Likewise, angle ACT = 180 - 60 - 45 = 75° = ATC
Then...in triangle ATC.......because ACT = ATC....then AT = AC = 24
Then....the area of ABC = (1/2)(AB)(AC)sin60 =
(1/2)(24sin105/sin45)(24)sin(60) =
(288)sin(105)sqrt(2)(sqrt(3)/ 2=
(144) sin(105)(sqrt(6) = about 340.7 units^2
[ The exact result is (216 + 72√3) units^6 ]
