Question

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Answer 1

Step-by-step explanation:

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In the given figure, ST is a straight line and ray QP stands on it.

In the given figure, ST is a straight line and ray QP stands on it.∴ ∠PQS + ∠PQR = 180º (Linear Pair)

In the given figure, ST is a straight line and ray QP stands on it.∴ ∠PQS + ∠PQR = 180º (Linear Pair)∠PQR = 180º − ∠PQS .........(1)

In the given figure, ST is a straight line and ray QP stands on it.∴ ∠PQS + ∠PQR = 180º (Linear Pair)∠PQR = 180º − ∠PQS .........(1)∠PRT + ∠PRQ = 180º (Linear Pair)

In the given figure, ST is a straight line and ray QP stands on it.∴ ∠PQS + ∠PQR = 180º (Linear Pair)∠PQR = 180º − ∠PQS .........(1)∠PRT + ∠PRQ = 180º (Linear Pair)∠PRQ = 180º − ∠PRT ...........(2)

In the given figure, ST is a straight line and ray QP stands on it.∴ ∠PQS + ∠PQR = 180º (Linear Pair)∠PQR = 180º − ∠PQS .........(1)∠PRT + ∠PRQ = 180º (Linear Pair)∠PRQ = 180º − ∠PRT ...........(2)It is given that ∠PQR = ∠PRQ.

In the given figure, ST is a straight line and ray QP stands on it.∴ ∠PQS + ∠PQR = 180º (Linear Pair)∠PQR = 180º − ∠PQS .........(1)∠PRT + ∠PRQ = 180º (Linear Pair)∠PRQ = 180º − ∠PRT ...........(2)It is given that ∠PQR = ∠PRQ.Equating equations (1) and (2), we obtain

In the given figure, ST is a straight line and ray QP stands on it.∴ ∠PQS + ∠PQR = 180º (Linear Pair)∠PQR = 180º − ∠PQS .........(1)∠PRT + ∠PRQ = 180º (Linear Pair)∠PRQ = 180º − ∠PRT ...........(2)It is given that ∠PQR = ∠PRQ.Equating equations (1) and (2), we obtain180º − ∠PQS = 180° − ∠PRT

In the given figure, ST is a straight line and ray QP stands on it.∴ ∠PQS + ∠PQR = 180º (Linear Pair)∠PQR = 180º − ∠PQS .........(1)∠PRT + ∠PRQ = 180º (Linear Pair)∠PRQ = 180º − ∠PRT ...........(2)It is given that ∠PQR = ∠PRQ.Equating equations (1) and (2), we obtain180º − ∠PQS = 180° − ∠PRT∠PQS = ∠PRT

In the given figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT Concept: Pairs of Angles.

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Answer 2

Answer:

"∠PQS = ∠PRT

Hence proved.

Given:

∠PQR = ∠PRQ

To prove:

∠PQS = ∠PRT

Proof:

∠PQR + ∠PQS  = 180° (by Linear Pair axiom)

∠PQS = 180° –  ∠PQR — (i)

∠PRQ + ∠PRT= 180° (by Linear Pair axiom)

∠PRT = 180° – ∠PRQ — (ii)

∠PRQ = 180° –  ∠PQR

[∠PQR = ∠PRQ]

From (i) and (ii), we can conclude the following,

∠PQS = ∠PRT= 180° – ∠PQR

∠PQS = ∠PRT

Hence, ∠PQS = ∠PRT

Hence proved the given terms."