In a circle triangle ABC is inscribed,AOC is diameter and x and y lies in between arc AB and BC respectively,if arc AXB=1/2arc BYC.Find angle BOC
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Figure:triangle ABC is such that AC passes through centre O and ANGLE ABC lie in semi circle.joinOB
Let Angle BOC be x
Therefore angle AOB= 180-x{linear pairs}
Length of arc AXB=1/2length of arcBYC
180-x/360×pier^2=1/2×x/360×pie×r^2
360-2x=x
360=3x
120=x=angleBOC
in triangle BOC
BO=CO{RADIUS}
angleOBC=angleOCB
Therefore by anlge sum property
2angleBCO+120=180
angleBOC=30
ansss:
Sorry its bit messy for any query about question do not hesitate to ask
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Figure:triangle ABC is such that AC passes through centre O and ANGLE ABC lie in semi circle.joinOB Let Angle BOC be x Therefore angle AOB= 180-x{linear pairs} Length of arc AXB=1/2length of arcBYC 180-x/360×pier^2=1/2×x/360×pie×r^2 360-2x=x 360=3x 120=x=angleBOC in triangle BOC BO=CO{RADIUS} angleOBC=angleOCB Therefore by anlge sum property 2angleBCO+120=180 angleBOC=30 plz mark tnx plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz ha ha
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