Question

The angle subtended by an arc at the centre is 45 degrees then the angle subtended by the arc at any point on the remaining part of the circle is​

Answer 1

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It is 1 radian, or 57.29 degrees.

Let's find out. Let's consider a circle having radius as r units.

The circumference subtends an angle of 360° or 2π radians at the center.

We'll take the ratio.

Here, x is the unknown angle subtendes by the arc of length equal to radius.

2πr÷r=2π÷x

x=1 radian.

Or in degrees

2πr÷r=360°÷x

x=360°÷2π

=360°÷6.28

=57.29 degrees.

Answer 2

Answer:

Consider a circle with centre “O”. Here the arc PQ of the circle subtends angle POQ at Centre O and ∠PAQ at a point A on the remaining part of the circle.

To prove: ∠POQ = 2∠PAQ.

To prove this, join AO and extend it to point B.

There are two general cases while proving this theorem.

Consider a triangle APO,

Here, OA = OP (Radii)

Since, the angles opposite to the equal sides are equal,

∠OPA = ∠OAP …(1)

Also, by using the exterior angle property (exterior angle is the sum of interior opposite angles),

We can write,

∠BOP = ∠OAP + ∠OPA

By using (1),

∠BOP = ∠OAP + ∠OAP

∠BOP = 2∠OAP… (2)

Similarly, consider another triangle AQO,

OA = OQ (Radii)

As the angles opposite to the equal sides are equal,

∠OQA = ∠OAQ … (3)

Similarly, by using the exterior angle property, we get

∠BOQ = ∠OAQ + ∠OQA

∠BOQ = ∠OAQ + ∠OAQ (using (3))

∠BOQ = 2∠OAQ …(4)

Adding (2) and (4) we get,

∠BOP + ∠BOQ = 2∠OAP + 2∠OAQ

∠POQ = 2(∠OAP + ∠OAQ)

To prove ∠POQ = 2∠PAQ for this case, we can follow the steps as same as for case (1). But while adding (2) and (4), we have to follow the below steps.

∠BOP + ∠BOQ = 2∠OAP + 2∠OAQ

Reflex angle ∠POQ = 2(∠OAP + ∠OAQ) (Since, PQ is a Major arc)

Reflex angle ∠POQ = 2∠PAQ.

hope it will help you