In triangle ABC, we have angle BAC = 60 degrees and angle ABC = 45 degrees. The bisector of angle A intersects line BC at point T, and AT = 24. What is the area of triangle ABC?
+128475 A
B C
Angle BAT = 30°
And....by the Law of Sines
AB / sin ATB = AT / sin ABC
AB / sin 105 = 24 / sin 45
AB = 24 sin (105) / sin 45 =
24 sin (60 + 45) / sin 45 =
24 [ sin (60)oos(45) / sin (45) + sin (45)cos (60) /sin (45) ] =
24 [ sin (60) + cos (60) ] =
24 [ sqrt (3) /2 + 1/2 ] =
12 [ sqrt (3) + 1 ]
And using the Law of Sines again
AB / sin ACB = BC / sin BAC
12 [ sqrt (3) + 1 ] / sin (75) = BC / sin (60)
12 [ sqrt (3) + 1 ] = BC sin (75) /sin (60)
12 [sqrt (3) + 1 ] * sin (60) = BC [ sin(30)cos (45) + sin (45)cos(30) ]
6 [ sqrt (3) + 1] * sqrt (3) = BC [ (1/2)(sqrt (2) / 2 + (sqrt (2) /2) sqrt (3) /2 ]
6 [ sqrt (3) + 1 ] * sqrt (3) = BC [ sqrt (2) + sqrt (6) ] / 4
[ 24 [ sqrt (3) + 1] *sqrt (3) ] / [ sqrt (2) + sqrt (6) ] = BC
[ 24 (3 + sqrt (3) ] / [ sqrt (2) + sqrt (6) ] = BC
[ 24 (sqrt (3) + 3 ] [ sqrt (6) - sqrt (2) ] / 4 = BC
[ 24 (sqrt (18) + 3sqrt (6) - sqrt(6) - 3sqrt (2) ] = BC
6 [ ( 3sqrt (2) + 2sqrt (6) - 3sqrt (2) ] = BC
12 sqrt (6) = BC
So...the area of ABC =
(1/2)AB * BC * sin (45) =
(1/2) [ 12 (sqrt (3) + 1) ] [ 12 sqrt (6) ] sqrt (2) / 2 =
(1/4) (144) (sqrt (3) + 1) (sqrt (12) ) =
36 [ sqrt (36) + 2sqrt(3) ] =
36 [ 6 + 2sqrt (3) ] =
36 (2 * 3 + 2 sqrt (3) ] =
72 [ 3 + sqrt (3) ] units^2 ≈ 340.7 units^2
