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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
avatar+109740 
+1

 

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

 

 

cool cool cool

 Feb 17, 2020
 #3
avatar+531 
0

I wish I could delete my "answer". I made a boo-boo. laugh

Dragan  Feb 18, 2020
edited by Dragan  Feb 18, 2020

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