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In triangle $$XYZ$$$$\angle X = 60^\circ$$ and $$\angle Y = 45^\circ$$ . Point $$D$$ lies on $$\overline{YZ}$$ such that $$\overline{DX}$$ bisects $$\angle ZXY$$ . If $$XD = 24,$$ then find the area of triangle $$XYZ$$.

Feb 17, 2020

#1
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Using the angle bisector theorem, the area of triangle XYZ works out to 108 + 144*sqrt(3).

Feb 17, 2020
#2
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Angle DXY = 30°

Angle XDY  =   105 °   and   angle DYX  = 45°   and angle  XZY  = 75°

Using the Law of  Sines

XY / sin XDY  =  XD / sin DYX

XY  / sin 105 = 24 / sin 45

XY =  24  sin (105) /sin (45)                   {sin105  = sin  75 }

XY =  24  sin 75   / sin 45

XY   = 24 [ sin (30 + 45) ]/ sin(45)

XY =  24 ( √2)  [ sin 30 cos 45 + sin 45 cos 30 ]

XY = 24√2 [ (1/2) (1/√2)  +  (1/√2) (√3/2]

XY = 24√2  [ ( 1 + √3) / [ 2√2]  =   12 (1 + √3)

And  using the Law of Sines again

XY /sin XZY  =  ZY / sin ZXY

12 ( 1 + √3) / sin 75  = ZY / sin 60

12 (1 + √3) /  [  ( 1 + √3) / 2√2) ]  = ZY / [√3/2]

ZY  = (24√2) (√3/2)  =  12√6

So....the area of triangle XYZ  =   (1/2) (XY) (ZY)sin 45  =  (1/2)  ( 12 [ 1 + √3 ]) (12√6)  * 1 / √2  =

72 ( 3 + √3)  units^2  ≈  340.7 units^2

Feb 17, 2020
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