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?
+128474 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
