can anyone explain me this math problem's solution

[bdiwakarteja]

[clock60]

until this problem, i thought that i am not too weak in geometry, but i am weak..
(2) as we know triangle BCO is isosceles (as BO=OC=radius)
and ABC also isosceles as AB=CO=BO
we know that angle BCO=40, so is angle CBO
and ABO+OBC=180, as CBO=40, ABO=180-40=140
and the sum of two left in the triangle ABC=180-140=40, the value of angle BAO=40/2=20
sufficient

(1) done, but more than an hour....
angle BAO=AOB=x,angle ABO=180-2x
angle OBC=BCO=y, angle BOC=180-2y

60+(180-2y)+x=180
2y-x=60

y+(180-2x)=180, y=2x, 2*2x-x=60, 3x=60, x=20, sufficient

[David@VeritasPrep]

you are correct, the answer is D.

[Ian Stewart]

Here is a better way to approach statement 1.

The measure of an angle is equal to the measure of the central angle (that cuts out the same arc) divided by the number of radii that the angle is away from the arc that it cuts out.

Let's take this particular problem, angle COD is a central angle. It has a measure of 60 degrees, since it is just one radii from the minor arc CD the measure of the angle is 60 degrees. Angle CAO is 3 radii away from arc CD - one diameter (which is two radii) plus another radii (beyond the circle). We know this because line segment AB is equal to line segment OC, which is a radius. Since angle BAO is 3 radii away we divide the central angle of 60 by 3 and get a measure of 20 degrees.

Unless I've completely misunderstood the above, this simply isn't true. If I understand correctly, you're claiming that the length of AD is equal to 3 radii, and that this implies that angle CAD is 1/3 of the central angle COD:

First, the line AD in this question is *not* the length of 3 radii; it's shorter than that. There are many ways to see this; for example, in triangle AOB, the sides AB and OB are both the length of a radius. By the triangle inequality, the sum of two sides exceeds the third: AB+OB > AO, or, using r for the radius, 2r > AO. Thus the length of AD = AO + OD < 2r + r, and the length of AD is thus less than 3r.

It certainly does not follow from the fact that AB is the length of a radius that AO is the length of two radii - indeed the fact that point B is *not* on the line AO guarantees that this *cannot* be true (since AOB thus forms a triangle with two sides of length r).

Further, the 'theorem' you are using, that would allow us to divide the central angle by AD/radius, is not true in general. It *will* be true for two points: the center of the circle, and the endpoint of a diameter. For any other point inside or outside a circle, this will never give you the correct angle. You can quite easily confirm this by taking any example point strictly outside of the circle and using the sine law.

Now, if I've misunderstood what you're trying to say, my apologies, and if so, I'd be interested to know what misinterpretation I've made.

_____

In any case, there is no need to memorize specialized angles theorems for the GMAT. For this question, you only need to know three elementary facts about angles:

* in an isosceles triangle, you have two equal angles, opposite the two equal sides
* the angles in a triangle add to 180
* the angles along a straight line add to 180

As long as you notice that each radius is equal in length, and so see that you have isosceles triangles, you can answer this question without any additional theorems.

[clock60]

hi guys
again to this problem. st 1 is solvable in 1 min if someone knows interesting property of angles within circle, now i know
the value of angle CAO=1/2*(arc CD-arc BE)
we are given that acr CD=60, and angle BAO=angle BOE as angles in the isosceles triangle ABC, say angles =x
x=1/2(60-x), 2x=60-x, 3x=60, x=20
that`s all

[silentuser]

clock60,

which point is E?

[clock60]

clock60,

which point is E?

hi sulentuser
sorry for not pointing out
point E lies on the line between A and D, it is arc BE