If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then ∠BAO =

(a) 60°

(b) 45°

(c) 30°

(d) 15°

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Solution

We will associate the given information in the following figure.

Since AO = r (radius of circle)

AM = (given)

Extended OM to D where MD =

Consider the triangles AOM and triangle AMD

So by SSS property

So AD = AO = r and OD=OM+MD=r

Hence ΔAOD is equilateral triangle

So

We know that in equilateral triangle altitudes divide the vertex angles

Hence option (c) is correct.

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