smyarga wrote:
desaichinmay22 wrote:
smyarga wrote:
Since, CD and AB are two diameters of the circle, so their intersection is the centre. Let's denote this centre O. Angle CEB is inscribed, and angle COB is corresponding central angle (arc CAB). The value of angle COB=180+48=228 degrees. The measure of inscribed angle is always half the measure of the central angle. Angle CEB is equal =228/2=114 degrees.
The correct answer is A.
Hi,
I thought angle COB is 180-42=138 degrees. You have derived angle COB of 228 degrees.
Can you please explain.
That's why I mentioned arc CAB. You don't really need to understand the notions of major and minor arc. Corresponding central angle means that it stays on the same arc as your inscribed angle.
another approach,
join C and A. since AB is the diameter, A, C, E and B form a cyclic quadrilateral. now opposit angles of cyclic quadrilaterals sum up to 180 degrees. considering the point of intersection of the 2 diameters as O.
angle AOC = 48 degrees. ........... given
OA = OC, radius of the circle, hence trianlge OAC is isoceles.
thus angle ACO = angle OAC = 66 degrees.
now,
angle OAC + angle CEB = 180 degrees......... since opposit angles of cyclic quadrilaterals sum up to 180 degrees.
therefore, angle CEB = 180 - 66 = 114 degrees.
hope this was helpful.
kudos if you like.