Math, asked by farhan96125, 11 months ago

the angle subtended by the major segment of a circle is an obtuse angle( True or False)​

Answers

Answered by arihantasharma666
3

Answer:

false

Step-by-step explanation:

Sol:

Given : A minor segment AB of a circle and two points C and D in BA and AB. respectively.

To prove : angle ACB is acute and angle ADB is obtuse.

Construction : Join OA and OB

Proof :

we know that angle subtended by an arc of a circle at its centre is twice the angle subtended by it at any point of the alternate segment of circle.

Since AB is minor arc and angle ACB is the angle formed by it in alternate segment.

Therefore 2∠ACB = ∠AOB.

=> 2 ∠ACB  < 180° ( since ∠AOB is an angle of triangle AOB)

=> ∠ACB < 90°

=> ∠ACB is an acute angle.

Now BA is the major arc and angle ADB is the angle formed by it in the alternate segment

Therefore 2∠ADB = m ( arc BA)

=>  2∠ADB = 360 - m ( arc AB)

=> 2∠ADB = 360 - ∠AOB  ( since  ( arc AB) = angle AOB )

=>  2∠ADB > 360 - 180  ( since ∠AOB < 180 )

=>  ∠ADB > 90°

=> ∠ADB is obtuse.

Answered by harinimusique20
0

Answer:

FALSE

Step-by-step explanation:

Given : A minor segment AB of a circle and two points C and D in BA and AB. respectively.

To prove : angle ACB is acute and angle ADB is obtuse.

Construction : Join OA and OB

Proof :

we know that angle subtended by an arc of a circle at its centre is twice the angle subtended by it at any point of the alternate segment of circle.

Now BA is the major arc and angle ADB is the angle formed by it in the alternate segment

Therefore 2∠ADB = m ( arc BA)

=>  2∠ADB = 360 - m ( arc AB)

=> 2∠ADB = 360 - ∠AOB  ( since  ( arc AB) = angle AOB )

=>  2∠ADB > 360 - 180  ( since ∠AOB < 180 )

=>  ∠ADB > 90°

=> ∠ADB is obtuse.

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