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Question

A chord AB subtends 120° at the center O of a circle whose radius is 10cm Find (i) the length of the minor arc AB (ii) the area of the triangle AOB (iii) the area of the major segment cut off by chord AB.

Answer 1

(i) arc AB/20(pi)=120/360=1/3, so AB=20(pi)/3 (one third of the circumference)=20.94cm.

(ii) draw perpendicular from O to chord AB at C to make two back-to-back triangles of equal area with OC as the common side. Angle AOC=BOC=60 degrees. AC/OA=sin60=sqrt(3)/2, AC=5sqrt(3), because OA=10. OC=OAcos60=10/2=5. Triangle OCB is congruent to OCA, so they have the same area=(1/2)AC*OC=(1/2)25sqrt(3). Therefore the area of AOB=25sqrt(3)=43.3 sq cm.

(iii) area of major segment AB/area of circle=(360-120)/360=240/360=2/3, therefore area=200(pi)/3=209.44 sq cm.