Question

Co=120
(2) In the figure. O is the centre of the circle.
mLAOB = 75°, then find
(1) mare AXB)
(ii) m(are AYB).
Solution :​

Answer 1

Answer:

m(arc AXB)= mLAOB = 75. (central angle)

m(arcAYB)=360-m(arc AXB)

360-75

285

^m(arcAYB)=285

360-