In figure, `angleACB=40^(@) ." Find "angleOAB`.

Explanation

Given, `angleACB=40^(@)`
We know that, a segment subtends an angle to the circle is half the angle subtends to the centre.
`:. angleAOB=2angleACB`
`rArr angleACB=(angleAOB)/2`
`rArr 40^(@)=1/2angleAOB`
`rArr angleAOB=80^(@)` ….(i)
In `DeltaAOB, AO=BO` [both are the radius of a circle]
`rArr angleOBA=angleOAB` ....(ii)
[angles opposite to the equal sides are equal]
We know that, the sum of all three angles in a triangle AOB is `180^(@)`.
`:. angleAOB+angleOBA+angleOAB=180^(@)`
`rArr 80^(@)+angleOAB+angleOAB=180^(@)` [form Eqs. (i) and (ii)]
`rArr 2 angleOAB=180^(@)-80^(@)`
`rArr 2angleOAB=100^(@)`
`:. angleOAB=100^(@)/2=50^(@)`