In the given figure, find angle BOC . D (x+10)^ с (x + 20) ^ 0 X ^ 0 A B
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Answer:
We know that the sum of all the angles in a straight line is 180
o
∠BOC+∠COD+∠AOD=180
o
.
x+20+x+x+10=180
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3x+30=180
o
3x=150
o
x=50
o
∠COD=50
o
∠AOD=x+10
o
=60
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∠BOC=x+20
o
=70
o
∴ ∠COD=50
o
, ∠AOD=60
o
and ∠BOC=70
o
.
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