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In quadrilateral ABCD, it is given that \(\angle A = 120^{\circ}\), angles B and D are right angles, AB = 13, and AD = 46. Then what is AC?

 Jan 14, 2021
 #1
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See the pic here :

 

 

Let A  =  (0,0)     D =  (0,46)

 

B=  (13 cos (-30)  , 13sin (-30)  )  =   (13sqrt (3)/2 , -13/2)  =  ( 6.5 sqrt (3)  , -6.5)

 

The  slope beteen  A and  B  =   (-6.5)/(6.5 sqrt (3) )  =   -1/sqrt (3)

 

A perpendicular line   to this  segment trhrough B  has the equation

 

y= sqrt (3)  ( x -  6.5 sqrt (3) )  - 6.5

 

y = sqrt (3) x  - 26

 

The x coordinate of the  intersection  of  this  line with the line y=  46   can be found as

 

sqrt (3)x  -26 =  46

 

sqrt (3)x  =  72

 

x =  72/sqrt (3)   =   24sqrt (3)  =  sqrt (1728)

 

C = ( 24sqrt (3) , 46 )

 

So......AC =    sqrt  [  (sqrt (1728))^2  + 46^2  ]   =   sqrt [1728 + 2116 ] = sqrt  (3844)   =   62  

 

 

cool cool cool

 Jan 15, 2021

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