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In quadrilateral $ABCD,$  $\angle A=75^\circ,\angle B=90^\circ,\angle C=105^\circ,$ and  $AB=BC=3\sqrt 2.$ Find $CD.$

 
 Nov 21, 2020
 #1
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First let's look at  △ABC.  By the Pythagorean Theorem,

 

AC2   =   AB2 + BC2   =   ( 3√2 )2 + ( 3√2 )2   =   18 + 18   =   36    and so    AC  =  6

 

△ABC  is a right isosceles triangle which means the base angles are both  45°

 

And so   m∠CAD  =  m∠BAD - m∠BAC   =   75° - 45°   =   30°

 

Now we can see that  △ACD  is a  30-60-90 triangle,

 

and in all 30-60-90 triangles, the side across from the 30° angle is half the hypotenuse.

 

In this case, that means   CD  =  1/2 * AC   =   1/2 * 6   =   3

 
 Nov 21, 2020

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