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Question
AB and AC are tangents drawn from a point A to a circle with centre O. If ∠BAC = 65°, then find the measure of ∠BOC.
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Solution

\[ \begin{array}{r l} \textbf{Given:} & \text{Tangents } AB \text{ and } AC \text{ drawn from an external point } A \text{ to a circle with centre } O, \text{ with } \angle BAC = 65^\circ. \\[4pt] \textbf{To Find:} & \text{The measure of } \angle BOC. \\[4pt] \textbf{Solution:} & \text{The radius through the point of contact is perpendicular to the tangent, hence } \angle OBA = \angle OCA = 90^\circ. \\[4pt] & \text{In the quadrilateral } ABOC, \text{ by the angle sum property of a quadrilateral:} \\[4pt] & \begin{aligned} \angle BAC + \angle OBA + \angle BOC + \angle OCA &= 360^\circ \\[4pt] 65^\circ + 90^\circ + \angle BOC + 90^\circ &= 360^\circ \\[4pt] \angle BOC &= 360^\circ - 245^\circ \\[4pt] &= 115^\circ \end{aligned} \\[4pt] \textbf{Answer:} & \angle BOC = 115^\circ. \end{array} \]
