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# help

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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?

Jan 27, 2019
edited by Guest  Jan 27, 2019

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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

Jan 27, 2019