akshatgupta87 wrote:Q.)The length of arc AXB is twice the length of arc BZC, and the length of arc AYC is three times the length of arc AXB. What is the measure of angle BCA?
A)20
B)40
C)60
D)80
E)120

Someone explain..

Thanks,
~Akshat

"The length of arc AXB is twice the length of arc BZC"

This implies angle BCA = 2 angle BAC

or BCA/360 2TTr = 2 BAC/360 2TTr

i.e angle BAC = 1/2 BCA

"the length of arc AYC is three times the length of arc AXB"

angle ABC = 3 BCA

call BCA x

x + 1/2x +3x = 180

x or BCA = 40

angle subtended at the center are in 1:2:6 ratio.

thus 9k = 360 giving k = 40
thus AXB subtends 80 deg at the center = 2* angle BCA

BCA = 40 deg.

Sorry for bumping into old thread. Can someone please let me know where am I going wrong with this?

axb=2(bzc)
ayc=3(axb)
ayc=6(bzc)

9bzc=360
bzc=40

bca=bzc+cya
bca=40+240
bca=280

struggled a lot.. things don't seem to get into this hard matter.

[delete post]

See attachement plz.
we have 9n=360, so n=40
as AOB=2ACB so m=40
answer will be B

The posted problem is virtually the same as the following:

The length of minor arc AB is twice the length of minor arc BC and the length of minor arc AC is three times the length of minor arc AB. What is the measure of angle BCA? (Please refer to the attachment)

1)20
2)30
3)40
4)50
5)60

An inscribed angle is formed by two chords.
Thus, angle BCA is an inscribed angle.

AB is the arc intercepted by inscribed angle BCA.
The degree measurement of an inscribed angle = 1/2 * the degree measurement of the arc intercepted by the inscribed angle.
Thus, angle BCA = (1/2)(arc AB).

Let BC = 1.
Since AB is twice BC, AB = 2.
Since AC is three times AB, AC = 3*2 = 6.
Thus, AB/circumference = 2/(1+2+6) = 2/9.

Since the circle = 360 degrees, the degree measurement of AB = (2/9)360 = 80.
Thus, angle BCA = (1/2)(arc AB) = (1/2)80 = 40.

The correct answer is C.