nevermind. I got it. I feel like an idiot!!

Here is the official solution. Sorry, I should have posted it. Someone pm-ed me. I should be more considerate for the others studying.

Also look at the attachment
-----------

According to the question, the ratio of the lengths of arc AC to arc AB to arc BC is 6: 2: 1. An inscribed angle cuts off an arc that is twice its measure in degrees. For example, if angle ACB is 30°, minor arc AB is 60° (or 60/360 = 1/6 of the circumference of the circle).

The angles of triangle ABC are all inscribed angles of the circle, so we can deduce the ratio of the angles of triangle ABC from the ratio of the lengths of minor arcs AC, AB and BC. The ratio of the angles will be the same as the ratios of the arcs.

The ratio of angle ABC to angle BCA to angle BAC is also 6: 2: 1. Using the unknown multiplier strategy, whereby a variable can be multiplied by each segment of a ratio to define the actual values described by that ratio, we might say that the interior angles equal 6x: 2x: x.
The sum of the angles in a triangle is 180, so x + 2x + 6x = 180.
9x = 180
x = 20
Angle BCA = 2x or 40

The correct answer is B.

thanks for the detailed solution!

Bc=x
Ab=2x
Ac=3x

Since arcs, 9x=360

x=40

Since area of triangle is half of circle, x=20

Bca=40

The length of minor arc AB is twice the length of minor arc BC and the length of major arc AC is three times the length of minor arc AB.

I'm so confused here. BC = x, AB = 2x, the length of AC = 3(AB) = 6x!. However BC + AB = AC and BC+AB = 3x. How can 3x = 6x?

What am I doing wrong here? It's driving me crazy. Thanks!

ns88 wrote:Bc=x
Ab=2x
Ac=3x

Since arcs, 9x=360

x=40

Since area of triangle is half of circle, x=20

Bca=40

I figured it out. I was tricked by the picture into thinking that the major arc went clockwise. Major ARC ac should be going counter clockwise instead.

ns88 wrote:Bc=x
Ab=2x
Ac=3x

Since arcs, 9x=360

x=40

Since area of triangle is half of circle, x=20

Bca=40

Good way of solving. In fact I did this in the same way. But just to clarify, it is not the area, but the angle subtended at the center of the circle

Hope this Helps

I'm confused by this problem because when it refers to major arc AC, I thought it was referring to arc ABC. How is it possible that the smaller arc AC is the minor arc? I saw that someone in the forum wrote that it goes counterclockwise, but I'm not sure if that's the rule....

Can someone please explain?

let length of bc be x
length of ab is 2x
and therefore length of ac is 6x
add them up u get circumfrence therefore 9x = 2pi r
x = 2pir/9
we get the length of bc as 2pir/9
similarly we get length of ab as 4 pi /9
and ac as 12pi/9
find the angle subtended at the canter by each of these cords
we get angle by ab as 80 degrees
bc as 40
and ac as 240 degrees
now with center o OA = Oc
so angle ACO = 30
similarly find BCO = 70
subtract aco from BCO to get BCA as 40

Folks,

Sorry to resurrect this thread from the dead but I have a question.

How is it that you cannot apply the principle of the "ratio of the sides of a triangle = ratio of the corresponding angles" to right angle triangles such as the 30-60-90 (1-sqrt(3)-2) triangle? The solution to this question uses the ratios of the sides to deduce and calculate the angles by stating that the ratios are 6:2:1, therefore the angles are 120:40:20

Do another set of rules govern right-triangles? Do you see what I mean?

In the 30-60-90 right triangle, the angles 30-60 are in the ratio 1:2 and we know that the ratio of 1:sqrt(3) definitely isn't 1:2

Hope I'm not superconfusing myself here but some insight into this would help me a whole lot.

Thanks

Received a PM.

Actually, you are confusing something: the ratio of 6:2:1 does not refer to lengths of sides, it refers to lengths of arcs. That's an entirely different thing, since arcs have much more "in common" with angles than with lengths of sides of triangles.

So basically even if the ratio between the angles of a triangle is 3:2:1, that doesn't mean that this ratio applies to the sides (actually, it won't).