In the given figure, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°; Find:
(i) ∠ACB, (ii) ∠OBC, (iii) ∠OAB, (iv) ∠CBA.
(i) ∠ACB, (ii) ∠OBC, (iii) ∠OAB, (iv) ∠CBA.
Exercise 17 (A) | Q 8 | Page 258
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Here, $\angle A C B=\frac{1}{2} \operatorname{Reflex}(\angle AOB )=\frac{1}{2}\left(360^{\circ}-140^{\circ}\right)=110^{\circ}$
(Angle at the centre is double the angle at the circumference subtended by the same chord)
Now, OA = OB (Radii of same circle)
$\therefore \angle OBA =\angle OAB =\frac{180^{\circ}-140^{\circ}}{2}=20^{\circ}$
∴ ∠CAB = 50° - 20°= 30°
ΔCAB,
∠CBA - 180° -110°- 30° = 40°
∴ ∠OBC = ∠CBA + ∠OBA = 40° + 20° = 60°
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