In the given figure, O is the centre of a circle in which ∠AOC = 100°. Side AB of quad. OABC has been produced to D. Then, ∠CBD = ?
(a) 50°
(b) 40°
(c) 25°
(d) 80°

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Solution

(a) 50°
Take a point E on the remaining part of the circumference.
Join AE and CE.

Then $∠ AEC = \frac{1}{2} ∠ AOC = (\frac{1}{2} \times 100 °) = 50 °$
Now, side AB of the cyclic quadrilateral ABCE has been produced to D.
∴ Exterior ∠CBD = ∠AEC = 50°

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In the given figure, O is the centre of a circle in which ∠AOC = 100°. Side AB of quad. OABC has been produced to D. Then, ∠CBD = ? (a) 50° (b) 40° (c) 25° (d) 80°

(a) 50° Take a point E on the remaining part of the circumference. Join AE and CE. Then ∠AEC=12∠AOC=12×100°=50° Now, side AB of the cyclic quadrilateral ABCE has been produced to D. ∴ Exterior ∠CBD = ∠AEC = 50°

In the given figure, O is the centre of a circle in which ∠AOC = 100°. Side AB of quad. OABC has been produced to D. Then, ∠CBD = ?
(a) 50°
(b) 40°
(c) 25°
(d) 80°

(a) 50°
Take a point E on the remaining part of the circumference.
Join AE and CE.

Then $∠ AEC = \frac{1}{2} ∠ AOC = (\frac{1}{2} \times 100 °) = 50 °$
Now, side AB of the cyclic quadrilateral ABCE has been produced to D.
∴ Exterior ∠CBD = ∠AEC = 50°