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ABCD is a parallelogram such that \( AB=\sqrt{3}\cdot BC.\) Diagonals   \( \overline{AC}\) and \(\overline{BD}\)  intersect at \( X \) and \(\angle ABC=\angle BXC.\)Find the measure of \(\angle{ACB}\) in degrees.

 

~~~Thanks~~~

 Dec 8, 2019
edited by RandomPerson  Dec 9, 2019
edited by RandomPerson  Dec 9, 2019
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Angle ACB  =  Angle XCB

Angle ABC  = Angle BXC

 

So.....by AA congruency......triangle CBA  is similar to triangle  CXB

 

Therefore

 

AB/ BC  =√3/ 1  = BX / XC ⇒  BX = √3 *  XC   ⇒   (BX)^2  = 3(CX)^2

 

And  by a property of  parallelograms,   AC  = 2CX

 

And

 

CX/ BC  =  BC / CA

 

BC^2  =  CA * CX

BC^2  =  2XC * CX

BC^2  = 2(CX)^2

BC  = √2 * CX

 

 

So .....using the Law of Cosines

 

BC^2  =  BX^2  + CX^2  -  2(BX)(CX) cos (CXB)

 

2(CX)^2  = 3(CX)^2 + (CX)^2  - 2(√3*CX * CX)  cos (CXB)

 

-2(CX)^2  =    -  2√3 * (CX)^2 cos (CXB)

 

1 / √3  =  cos (CXB)

 

So  the sin  of  CXB   =    √ [ 3 - 1]/ √3   =    √(2/3)

 

So....by the Law of Sines

 

sin XCB  / BX  =  sin (CXB) / BC

 

sin XCB / (√3 CX)  =  √(2/3) / (√2 CX)

 

sin XCB =   √3 * (√2/ √3)

                    _________     =      1

                           √2

 

sin (XCB)  =  1  = sin ACB   

 

arcsin (1)  =  ACB  =  90°

 

 

cool cool cool

 Dec 9, 2019
edited by CPhill  Dec 9, 2019

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