1.

In the adjoining figure, BP is the diameter of the circle. If `angle ABP=60^(@)`, then find `angle AQB`.

Answer» BP is the diameter of the circle.
`thereforeangle BAP=90^@`
In `Delta ABP`,
`angle APB=180^@-(angle BAP+angleABP)`
`=180^@-(90^@+60^@)`
`30^@` (angles of same segment )
Now, `angle AQB=angle APB`
`rArr angle AQB=30^@`


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