In the given figure, AB = BC = CD and ∠ABC = 132° . Calcualte: ∠AEB
Exercise 17 (B) | Q 7.1 | Page 265
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In the figure, O is the centre of circle, with AB = BC = CD.
Also, ∠ABC = 132°
In cyclic quadrilateral ABCE
∠ABC + ∠AEC = 180° (sum of opposite angles)
→ ∠132 + ∠AEC = 180°
→ ∠AEC = 180° -132°
→ ∠AEC = 48°
Since, AB = BC,∠AEB = ∠BEC (equal chords subtends equal angles)
$\therefore \angle AEB =\frac{1}{2} \angle AEC$
$=\frac{1}{2} \times 48^{\circ}$
$=24^{\circ}$
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