The length of minor arc AB

[bgussin]

Can someone help me with the following question? It seems to me there must be a typo here, the explanation given by the test only works if they are referring to MINOR arc AC. I must be missing something, please help if you can.

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. What is the measure of angle BCA?

[bgussin]

[vikaspanwar229]

Dear bgussin,
It is a good geometry prob. thanks.........
Ans is "40".

[bgussin]

Could you elaborate on how you came up with that solution. I dont understand how they are referencing the "major" arc in solving this problem.

[bgussin]

Would some one be willing to give me a formal solution to this problem? I am taking the test on Sat, so I would like to know more than "It is a good geometry prob...here's the answer". Please note I pasted the picture found in the original problem in the second reply to this thread.

[vikaspanwar229]

Let Arc BC = x, So Arb AB = 2x & AC=6x; right.
Now circum of circle will be = 9x, angle at the centre by circum = 360
so x=360/9=40
Now Meas of angle BCA, formed by arc AB on circum.
will be half of the angle formed by AB arc at centre.
Angle at centre BOA=2x=80,
So BCA=40

Thanks
Vikas

[eshieh06]

I agree with bgussin -- I'm also confused as to how major arc AC can be 6 times the length of minor arc AB. If BC is x, then AB is 2x, and major arc AC is 3x (2x + x).

I also don't understand how this info translates to knowing what angle BCA is. Can someone elaborate?

Thanks for your help.

[RonPurewal]

hmm ok, here's the story here.

it turns out that the picture that's been posted is grossly not to scale, and that the arc that's drawn directly from a to c (the arc at the bottom of the picture - not the arc that threads through point b) is the arc that's being referred to.

HERE'S THE CONFUSION (which is certainly the fault of the problem author(s)):
what they're TRYING to do is to call the arc "AC" in order to tell you that they're NOT talking about the arc that goes through point b.
then they're ALSO trying to inform you that this is a major arc.

this isn't very good writing. what they should do is just insert another point between a and c - call it "Q" - and then just talk about "arc AQC". there shouldn't be any need to mention major/minor, as the algebra will reveal that automatically (and it's not good form to drop needless hints).

also, technically, they HAVE to do this, because major arcs must be named by three points. if we use two letters to name the arc, then that defaults to the minor arc, a situation that will actually destroy the integrity of this problem (since that arc would actually be arc ABC here).

i will submit this problem for revision.

--

in any case:

imagine that there's point Q between a and c, and then replace "major arc AC" with "arc AQC" in the problem statement.

then:

according to the two given facts, arc BC is the smallest, so let's call it 'x'.
then according to the first statement, arc AB is '2x'.
then according to the second statement, arc AQC is '6x' (since it's three times 2x).

together these form a complete circle, so x + 2x + 6x = 360, or, x = 40.
therefore, arc ab is 80.
the desired angle is half the measure of this arc (the arc that it "intercepts"), so it's 40.

[ronak_svit]

Hi Ron,
I have a doubt about your equation :
Shouldnt it be x + 2x + 6x = 2*pi*r ( instead of 360 ??)
Here we are adding the lenght of arcs and not degrees...though we shall get the same answer at the end as 40 as follows:

x = (2/9)(pi*r)

For AB we need to get 2x which is (4/9)(pi*r)

(4/9)(pi*r) = (Angle in question /360 ) * (2*pi*r)
Angle in question = 40

[Kweku.Amoako]

AB = 2BC
AC = 3BC

OR AC = 3AB = 6BC

some imagination is reuired here. if you redraw the diagram such that you have a line from the center of the circle(call it point P) to each point (A,B and C) then

AB is equivalent to APB
BC is equivalent to BPC
AC is equivalent to APC

notice that on this same diagram, if you insert the orignal lines in the diagram which are lines AB, BC and AC you will realize that

BCA = APB / 2

so if we can find APB then we can find BCA

now we already know that AC = 3AB = 6BC which is equivalent to APC = 3APB = 6 BPC. We also know that from circle theory that APB + BPC +APC = 360 . Therefore using ratios if we make AB = x then AC = 3x and BC = 0.5x

x + 3x + 0.5 x = 360

4.5x = 360

x = 80
pugging this back into the equation BCA = APB / 2 since we found AB = APB = 80

then BCA = 80 /2 = 40

[Ben Ku]

I have a doubt about your equation :
Shouldnt it be x + 2x + 6x = 2*pi*r ( instead of 360 ??)
Here we are adding the lenght of arcs and not degrees...

Ronak: You are technically correct, because we are adding the lengths of the arcs. We do get the same answer because measures of arcs on a circle does not depend on the size of the circle (i.e. radius). Eventually we'll get the same degree measures because the "r"s cancel out.

[ogbeni]

Help!

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

Does 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

[Ben Ku]

Help!

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

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

The principle you stated is NOT true. The ratios of sides are NOT the ratios of angles. You should NEVER apply this statement to ANY triangle.

I think you might be confusing this with another geometry principle:
In a triangle, the largest angle is opposite the longest side, and the smallest angle is opposite the shortest side. However, this principle does NOT give us any information relating the ratios of sides or angles.