please solve this...........please
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In ∆COD All sides are equal
Hence, it is equilateral triangle
Join CB
So, angle ACB And Angle ECB=90 (Angle in semicircle)
And, ANGL COD=ANGL COD=ANGL COD=60
ANGLE CBE=1/2 angle COD( angle made by arc CD at center and the circumference)
ANGL CBE=30
Therefore, ANGL COD= 90+30+X=180
X=60
Hence, it is equilateral triangle
Join CB
So, angle ACB And Angle ECB=90 (Angle in semicircle)
And, ANGL COD=ANGL COD=ANGL COD=60
ANGLE CBE=1/2 angle COD( angle made by arc CD at center and the circumference)
ANGL CBE=30
Therefore, ANGL COD= 90+30+X=180
X=60
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