1.

In the adjoining , `O` is the centre of the circle. `ACB` is a segment. If `angle OAB=30^(@)`, then find the value of `angle ACB`.

Answer» In `Delta OAB`,
`AO=BO` (radii of circle)
`rArrangle ABO=angleBAO`
`rArrangle ABO=30^@`
Now, `angle AOB=180^@-(angleABO+angle BAO)`
`=180^@-(30^@+30^@)`
`120^@`
Arc AB subtends angles at
centre `=angle AOB`
remaining circle `=angle ACB`
`therefore angle ACB=(1)/(2)angle AOB`
`=(1)/(2)xx120^@`
`=60^@`


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