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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