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In triangle $XYZ$, and . Point $D$ lies on such that bisects ​ If $XD = 24,$ then find the area of triangle $XYZ$.

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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 $XYZ$.

Mar 31, 2020

#1
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Z   75

D

60  X                           Y  45

We  can  use  the Law  of Sines  twice

XY /  sin XDY   =  XD /sin DYX

XY / sin (105)  =  24 / sin (45)

XY  =  24 sin (105) / sin (45)

XZ / sin DXY   =  24/ sin XZD

XZ / sin (30)  = 24 / sin (75)

XZ =  sin (30)  sin (105)

Area XYZ   =   (1/2) XY * XZ  sin (60)  =  (1/2) [ 24 sin (105) / sin(45) ]  [ 24 sin (30) / sin (105) ] * sin (60)  =

(1/2)  ( 24)^2  *  [ sin (30) * sin (60)  / sin (45)  ]  =

(1/2)  (24)^2  * [ √3/4]  / [ √2/2]   =

(1/2) (24)^2 * [ 2√3] / [ 4√2]  =

(1/2) (24)^2 * [ √[3/2] /2 =

(1/4) (576) * [ √3] /  [ √2]  =

144 √3 * √2  /  2  =

72√6  units^2

Mar 31, 2020
#2
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Using the angle bisector theorem, the area of triangle XYZ works out to 72 + 120*sqrt(3).

Mar 31, 2020