Question : Chords AC and BD of a circle with centre O, intersect at right angles at E. If $\angle OAB=25^{\circ}$, then the value of $\angle EBC$ is:
Option 1: $30^{\circ}$
Option 2: $25^{\circ}$
Option 3: $20^{\circ}$
Option 4: $15^{\circ}$
Correct Answer: $25^{\circ}$
Solution : Construction: Join $OB$ Using figure, $OA = OB =$ radius $\angle OAB = \angle OBA = 25^\circ$ In $\triangle AOB$ $\angle AOB + \angle OBA+\angle OBA = 180^\circ$ $\therefore \angle AOB = 180^\circ - 25^\circ- 25^\circ = 130^\circ$ To find: $\angle EBC$ Construction: Join $BC$ $\therefore$ $\angle BCA = \frac{\angle AOB}{2}$ (The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.) $= \frac{130^\circ}{2} = 65 ^\circ$ In $\triangle EBC$, $\angle EBC + \angle BEC+\angle BCE = 180^\circ$ $\therefore\angle EBC = 180^\circ-90^\circ - 65^\circ = 25^\circ$ Hence, the correct answer is $25^\circ$.
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