In figure A, B, and C are three points on a circle such that the angles subtended by the chords AB and AC at the centre O are 90 $^{\circ}$ and 110 $^{\circ}$ , respectively. Determine $∠$ BAC.

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Solution

OA= OB = Radius. (isosceles, triangle, base angles are equal.)
$∠$ OBA= $∠$ OMB = 45 $^{\circ}$
Similarly :
OC = OA = Radius (isosceles triangle, base angles are equal.)
$∠$ OAC = $∠$ OCA = $\frac{(180^{\circ} - 110^{\circ})}{2}$ = 35 $^{\circ}$ .
$∠$ BAC = $∠$ BAO + $∠$ OAC = 80 $^{\circ}$

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