Math, asked by JJMP, 3 months ago

can anyone answer these questions fully​

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Answered by Anonymous
4

Answer:

Theorem:

The angle subtended by an arc of a circle at its centre is twice of the angle it subtends anywhere on the circle’s circumference.

The proof of this theorem is quite simple, and uses the exterior angle theorem – an exterior angle of a triangle is equal to the sum of the opposite interior angles.

If the two opposite interior angles happen to be equal, then the exterior angle will be twice of any of the opposite interior angles.

Proof:

Consider the following figure, in which an arc (or segment) AB

A

B

subtends ∠AOB

A

O

B

at the centre O

O

, and ∠ACB

A

C

B

at a point C

C

on the circumference.

Angle Subtended by an Arc at the Centre of a circle

We have to prove that ∠AOB=2×∠ACB

A

O

B

=

2

×

A

C

B

. Draw the line through O

O

and C

C

, and let it intersect the circle again at D

D

, as shown.

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