+0  
 
0
99
2
avatar

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
avatar+111321 
+1

                  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

 

 

cool cool cool

 Mar 31, 2020
 #2
avatar
0

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

 Mar 31, 2020

22 Online Users

avatar