PS:

thumpin_termis · Posted: Thu Jul 05, 2007 2:55 pm Post subject: PS:

Answer is B. How do you solve this?

Guest · Posted: Thu Jul 05, 2007 3:10 pm Post subject:

The way I look at it...

AB=BC=AC

so it is an equilateral triangle (60 degrees)

Therefore, angle BAD is 120 degrees...

and since BA=AD

60/2 = 30

thumpin_termis · Posted: Thu Jul 05, 2007 3:40 pm Post subject:

jrbrown2 · Posted: Thu Jul 05, 2007 4:08 pm Post subject:

First, AB=AD b/c they are both the radius of the circle

From the question AB=BC=CD so AB=BC=CD=AD. From this we know that ABCD is a rhombus (all sides have to be equal)

In a rhombus the diagonals always bisect each other. BD bisects AC and vice versa.

Calling the point at which BD & AC bisect O, we know that AO is 1/2 AC (AC is the radius). So AO is also 1/2 AB because AB is also the radius.

Now looking at right triangle ABO we know that AB is twice AO. A right triangle w/ the hypotenuse twice a side proves that that's a 30-60-90 triangle we're looking at. 30 degrees is always opposite the shortest side so angle X is 30 degrees. Hence B