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 $XYZ$.
+128407 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
