+92 In quadrilateral $ABCD,$ $\angle A=75^\circ,\angle B=90^\circ,\angle C=105^\circ,$ and $AB=BC=3\sqrt 2.$ Find $CD.$
+9481 
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
