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Two chords AB and AC of a circle subtend angles equal to 90º and 150º, respectively at the centre. Find  \angleBAC, if AB and AC lie on the opposite sides of the centre.

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Given: Two chords AB and AC of a circle subtend angles equal to 90º and 150º, respectively at the centre.

AB and AC lie on opposite sides of the centre

In \triangleBOA        

OB = OA (Both equal to radius)

\angleOAB = \angleOBA … (i) (angles opposite to equal sides in a triangle are equal)

In \triangleOAB,

\angleOBA + \angleBAO + \angleAOB = 180°  (angle sum property of a triangle)

\Rightarrow \angleOAB + \angleOAB + 90° = 180°  [From equation (1)]

\Rightarrow 2\angleOAB = 180° – 90°

\Rightarrow \angleOAB = \frac{90^{\circ}}{2}= 45°

Now in \triangleAOC, AO = OC  (Both equal to radius)

\angleOCA = \angleOAC … (ii)    (angles opposite to equal sides in a triangle are equal)

Also,

\angleAOC + \angleOAC + \angleOCA = 180°  (angle sum property of a triangle)

150° + 2 \angleOAC = 180°                     

2\angleOAC = 180° – 150°

2\angleOAC = 30°

\angleOAC = 15°

\angleBAC = \angleOAB +\angleOAC = 45° + 15° = 60°

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