In the adjoining figure, chords AC and BD of a circle with centre O, intersect at right angles at E. If ∠OAB = 25°, calculate ∠EBC.

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Solution

OA = OB (Radii of a circle)
Thus, ∠OBA = ∠OAB = 25°
Join OB.

Now in ΔOAB, we have:
∠OAB + ∠OBA + ∠AOB = 180° (Angle sum property of a triangle)
$\Rightarrow$ 25° + 25° + ∠AOB = 180°
$\Rightarrow$ 50° + ∠AOB = 180°
$\Rightarrow$ ∠AOB = (180° – 50°) = 130°

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
i.e., ∠AOB = 2∠ACB
$\Rightarrow$ $∠ A C B = \frac{1}{2} ∠ A O B = (\frac{1}{2} \times 130 °) = 65 °$
Here,∠ACB = ∠ECB
∴ ∠ECB = 65° ...(i)

Considering the right angled ΔBEC, we have:
∠EBC + ∠BEC + ∠ECB = 180° (Angle sum property of a triangle)
$\Rightarrow$ ∠EBC + 90° + 65° = 180° [From(i)]
$\Rightarrow$ ∠EBC = (180° – 155°) = 25°
Hence, ∠EBC = 25°

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