plz sol both the questions
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4.) ∠AOB=∠AOD + ∠COD + ∠COB
( Given ∠BOC=40 , ∠COD=100
and since AB is a straight line , ∠AOB =180 )
180=∠AOD+100+40
∠AOD=40
In Triangle AOD, since all sides of triangle sum up to 180
180=∠AOD+∠ODA+∠DAO
180=40+∠ODA+∠DAO
( Since both sides OA and OD are equal because they are the radius of same circle
So Triangle AOD is an isosceles triangle
Therefore both angles opposite to equal sides ∠ODA and ∠DAO must be equal )
180=40+2*∠ODA
140=2*∠ODA
∠ODA=70
( Given ∠BOC=40 , ∠COD=100
and since AB is a straight line , ∠AOB =180 )
180=∠AOD+100+40
∠AOD=40
In Triangle AOD, since all sides of triangle sum up to 180
180=∠AOD+∠ODA+∠DAO
180=40+∠ODA+∠DAO
( Since both sides OA and OD are equal because they are the radius of same circle
So Triangle AOD is an isosceles triangle
Therefore both angles opposite to equal sides ∠ODA and ∠DAO must be equal )
180=40+2*∠ODA
140=2*∠ODA
∠ODA=70
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