+87 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 \(XYZ\).
Using the angle bisector theorem, the area of triangle XYZ works out to 108 + 144*sqrt(3).
+128408
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
