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Angle Subtended by an Arc of a Circle on the Circle and at the Center

In the figure...

Question

In the figure, O is the centre of the circle and the length of are AB is twice the length of are BC. If angle AOB = $108^{o}$ , find :

(i) $∠$ CAB,

(ii) $∠$ ADB.

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Solution

Given $\widehat{AB} = 2 \times \widehat{BC}, \angle AOB = 108^o$

(i) $\angle AOB = 2 \angle BOC$

$\Rightarrow \angle BOC = \frac{108}{2} = 54$

(ii) $\angle BAC = \frac{1}{2} \angle BOC$ (angle subtended at the center is twice that subtended at the circumference by a chord)

$\angle BAC = \frac{1}{2} \times 54 = 27^o$

(iii) Similarly, $\angle ACB = \frac{1}{2} \angle AOB = \frac{108}{2} = 54$

In cyclic quadrilateral $ADBC$

$\angle ADB + \angle ACB = 180^o$

$\Rightarrow \angle ADB = 180-54=126^o$

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Similar questions

In the adjoining figure A, D, B, C are four points on the circumference of a circle with centre O. Arc AB = 2 Arc BC and $∠$ AOB = $108^{0}$ . FInd:

i. $∠$ ACB

ii. $∠$ CAB

III. $∠$ ADB

∠ A O B = 70^{o}

△ O C D

O

A B

C D

5 : 3

A O B = 100^{\circ},

∠ C O D

a r c A B

a r c C D

The length of arc AB is twice the length of arc BC of a circle with centre O. If $∠$ AOB is $100^{o}$ then $∠$ BOC is

A B a n d B C

3 : 2

∠ A O B = 96^{o}

∠ B O C

∠ A B C

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