Question

in the circle with centre O and radius 6 cm if the length of a chord AB is 6 root 3 cm then angle AOB is
options
A. 110 degree
B.100 degree
c. 120 degree
d. 130 degree
​

Answer 1

Step-by-step explanation:

Given: In the circle with centre O and radius 6 cm if the length of a chord AB is 6 root 3 cm.

To find: Angle AOB is

options are:

A. 110 degree

B.100 degree

C. 120 degree

D. 130 degree

Solution:

Theorem used:

Perpendicular from centre of circle bisects chord.

Step 1: Construct the circle.

According to given dimensions construct the circle for better understanding.

Draw a perpendicular on chord AB from centre.

According to the theorem discussed above.

$AM =BM = 3 \sqrt{3} \: cm$

Step 2: Find the length of OM.

As AOM is right triangle.

Apply Pythagoras theorem to find OM.

${OA}^{2} = {OM}^{2} + {AM}^{2}$

${6}^{2} = ( {3 \sqrt{3} })^{2} + {OM}^{2}$

$36 = 27 + {OM}^{2}$

${OM}^{2} = 9$

$OM = 3 \: cm$

Step 3: Apply trigonometric ratio to find Angle AOM.

Let the angle AOM is $\theta$

$cos \theta = \frac{OM}{OA}$

$cos \theta = \frac{3}{6}$

$cos \theta = \frac{1}{2}$

$cos \theta = cos \: 60°$

or

$\bold{\theta = 60°}$

Step 4: Find angle BOM

∆AOM and ∆BOM are congruent by SSS criteria of congruency.

As

OA=OB [radius]

OM=OM [common]

AM=MB [By construction]

Thus,

Corresponding angles of both triangles are equal.

So,

$\bf \angle AOM= \angle BOM = 60°$

Step 5: Find $\angle AOB$

As

$\angle AOB=\angle AOM+\angle BOM$

$\angle AOB = 60° + 60°$

$\angle AOB=120°$

Final answer:

$\bf \angle AOB=120°$

Option C is correct.

Hope it will helps you.

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Answer 2

Answer:

120 degree

Step-by-step explanation:

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