Question

State and prove degree measure theorem.

Answer 1

ᴅᴇɢʀᴇᴇ ᴍᴇᴀsᴜʀᴇ ᴛʜᴇᴏʀᴇᴍ.

➩ states that angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the circle. ... In each of these three cases, the exterior angle of a triangle is equal to the sum of two interior opposite angles.

ʜᴏᴘᴇ Ɪᴛ ʜᴇʟᴘs ʏᴏᴜ Ғʀɴᴅ

Answer 2

⭐ Degree Measure Theorem

Degree Measure Theorem states that angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the circle. In each of these three cases, the exterior angle of a triangle is equal to the sum of two interior opposite angles.

➩ States that angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the circle.

Given :

An arc PQ of a circle C with center O and radius r with a point R in arc QP other than P or Q.To Prove : ∠POQ=2 ∠PRQ

Construction : Join RO and draw the ray ROM.

Proof :

There will be three cases as

(i) PQ⌢is a minor arc

(ii)PQ⌢ is a semi-circle

(iii)PQ⌢ is a major arc

Solution :

In each of these three cases, the exterior angle of a triangle is equal to the sum of two interior opposite angles.

Therefore, ∠POM=∠PRO+∠RPO ----- (1)

∠MOQ=∠ORQ+∠RQO. --- ----(2)

In △OPR and △O Q R

Now, OP = OR and OR = OQ (radii of the same circle)

∠PRO=∠RPO and ∠ORQ=∠R Q O (angles opposite to the equal sides are equal)

Hence, ∠POM=2∠PRO -------- (3)

And ∠MOQ=2∠ORQ ------ (4)

Case (1) : adding equations (3) and (4) we get

∠POM+∠MOQ=2∠PRO+2∠ORQ

∠POQ=2(∠PRO+∠ORQ)=2∠PRQ

∠POQ=2∠PR Q

Hence Proved.

Similarly, you can proceed for case (ii) and case (iii)

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