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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. - Mathematics

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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^@`

  Is there an error in this question or solution?
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APPEARS IN

 2009-2010 (March) (with solutions)
Question 6.2 | 3.00 marks
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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. Concept: Tangent to a Circle.
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