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