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Question
In the following figure, O is the centre of the circle and AB is a tangent to it at point B. ∠BDC = 65°. Find ∠BAO.
Solution
Form the given figure OB is the radius and therefore in triangle BDC
`∠DBC + ∠BDC + ∠BCD = 180^@`
`=> 90^@ + 65^@ + ∠BCD = 180^@`
=> ∠BCD = 25
Now, OE = OC = radius, ∠OEC = ∠OCE = 25° (as ∠OCE = ∠BCD)
=> ∠AED = ∠OEC = 25° (Vertically opposite angles)
Also, `∠ADE = 180^@ - 65^@ = 115^@`
Therefore in triangle AED
`∠BAO = 180^@ - 115^@ - 25^@ = 40^@`
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