In figure, DeltaO D C~ DeltaO B A ,/B O...

In figure, `DeltaO D C~ DeltaO B A ,/_B O C=125^o`and `/_C D O=70^o`. Find`/_D O C ,/_D C O and /_O A B`.

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We can apply lineat on line BD
`/_BOC + /_DOC = 180^0`
`/_DOC = 180^0-125^0`
`=55^0`
Now in triangle DCO,
`/_CDO + /_DCO + /_DOC = 180^0`
hence,
`/_DCO = 180^0-70^0-55^0`
...

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