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In the following figure, AB is the diameter of circle and CD is a chord equal to the radius. AC and BD when extended meet at E. ∠AEB = ?
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- 30°
- 60°
- 45°
- 90°
- 30°
Correct Option: B
As per the given figure in question ,
In ∆ OCD,
OC = OD = CD
∴ ∆ OCD is an equilateral triangle.
As we know that each angle of an equilateral triangle is 60° .
∴ ∠COD = 60°
Now, ∠CBD = ( 1/2 )∠COD
⇒ ∠CBD = 30°
∠ACB is angle of semi-circle.
∴ ∠ACB = 90°
⇒ ∠ BCE = 180° – ∠ACB = 180° – 90° = 90°
In ∆ BCE,
∠BCE = 90°, ∠CBE = ∠CBD = 30°
∴ ∠BCE + ∠CBE +∠CEB = 180°
⇒ 90° + 30° + ∠CEB = 180°
⇒ ∠CEB = 180° – 120° = 60°
⇒ ∠ AEB = 60°
