Math, asked by vrinda2882, 11 months ago

Prove degree measure theorem​


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Answers

Answered by sumitmazumdar
1

Ans. 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.

Given : An arc PQ of a circle C with center O and radius r with a point R in arc \(\stackrel\frown{QP}\) other than P or Q.To Prove : \(\angle POQ = 2\angle PRQ\)

Construction : Join RO and draw the ray ROM.

Proof : There will be three cases as

(i) \(\stackrel\frown{PQ}\)is a minor arc

(ii)\(\stackrel\frown{PQ}\) is a semi-circle

(iii)\(\stackrel\frown{PQ}\) is a major arc

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

Therefore, \(\angle POM = \angle PRO + \angle RPO \)  ............ (1) 

\(\angle MOQ = \angle ORQ + \angle RQO \).......... (2)  

\(In\space \triangle OPR \space and \space \triangle OQR \)

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

\(\angle PRO = \angle RPO\) and \(\angle ORQ = \angle RQO \)  (angles opposite to the equal sides are equal)

Hence, \(\angle POM = 2\angle PRO\)      .......... (3)

And \(\angle MOQ = 2 \angle ORQ\)     ............ (4)

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

\(\angle POM + \angle MOQ = 2\angle PRO + 2\angle ORQ \)

\(\angle POQ = 2(\angle PRO + \angle ORQ) = 2 \angle PRQ \)

\(\angle POQ = 2\angle PRQ\)

Hence Proved.

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

Answered by peehuthakur
1

Step-by-step explanation:

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.

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