Circle

This is an interesting problem. I am not satisfied with the explanation given in the solution you want to clarify.

From I all we know ACO=CAO which they must because OC =OA=radius. Beyond this we cannot conclude that ABO =AOC=130 because we don’t know what angles arcs AB and BC subtend with their respective central angles AOB and BOC. However we can say something about these angles. Draw line AC and you have Isosceles triangle ACO with angles ACO 25, CAO 25 and AOC 130. This comes from the info in I. Label the other portion of angle A as x and the other portion of angle B as y. You have no info about them. For example never assume they are 25. Don’t assume what you are not given or which you cannot prove conclusively. You now have triangle ABC with angles BAC =x, BCA= y, and ABC which we are to find. Since segments AB and AC form angle x, the arc this interior angle subtends must be twice x so Arc BC= 2x. Similarly BC and AC subtends 2y on arc AB. Since a central angle is equal to its arc, you know BOC = 2x and BOA= 2y and you already knew AOC = 130

So 2y + 2x= 130 from which it follows that x +y = 65. But x + y are the two angles in triangle ABC and if they sum to 65 then angle ABC is 180-65= 115 not 105 as in the posted solution. This is why I is sufficient. By merely drawing AC and radius OB, you should be able to derive all of these just by looking at the diagram. REMEMBER YOU ARE NOT TO FIND ABC, ONLY DETERMINE WHETHER IT CAN BE FOUND. ONCE YOU KNOW ALL THESE RELATIONSHIPS YOU CAN EASILY MAKE THAT DETERMINATION.

In II you are given AOC = 130 and this time you don’t know ACO=CAO=25 which was given in I. Frequent mistake is to use info from I in solving II which we will not do here. But you can still derive this information because AOC is still an isosceles triangle which means ACO=CAO=25 and the analysis above follows. II is also sufficient.

getso wrote:Hello ,

Can anybody explain the calculation below.

I have attached the question.

From statement 1, since OA=OC, angle ACO=angle CAO=25, then angle AOC150, then the angle
ABC=(360-150)/2=105. SUFF
From statement 2, ABC=(360-150)/2=105. SUFF

Angle subtended by the equal chords to a point on circle, are equal, but here OA and OC are radii, not chords, so let’s think differently to check the competency of statement (1). In fact, if we join A to C, we can find angle AOC and hence the angle ABC, with the same explanation as for (2) here under. So [spoiler]sufficient[/spoiler].

Statement (2) is also alone sufficient, see how:

If AC is a chord, then angle subtended by AC on the center (angle OAC = 130º) will be double of what it subtends in the segment that doesn’t contain B (it’s a theorem), let’s take D as a point on circle representing the alternate segment, then angle ADC will be 65º according to the theorem, and also ABCD would be a cyclic quadrilateral, hence angle ABC will be supplement of angle ADC (another theorem: opposite angles of a cyclic quadrilateral are supplementary) or angle ABC = 180º - 65º = 115º.

Mine [spoiler]D[/spoiler].

Skills like "The Alternate Segment Theorem in Circle" are tested under the range 700-800.

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https://www.mathsrevision.net/gcse/pages.php?page=13

From 1, we can say OAC = 25

Connecting O and B and considering OA and OB(both radius), we find
OBA = OAB = 25 + CAB -> 1

Similarly, Connecting O and B and considering OC and OB(both radius), we find
OBC = OCB = 25 + ACB -> 2

We know from the diagram and from 1 and 2 above,
ABC = OBA + OBC = 25 + CAB + 25 + ACB = 50 + CAB + ACB -> 3

We also,
ABC + CAB + ACB = 180 -> 4

So equating 4 and 3, we can get ABC

From 2, also we can find out OAC = 25 and similar above explanation holds.

It took me 3 minutes to do it first time, so worth spending this much time for a sum like this.