Question

in a triangle pqr ,side pq is produced to s so that qs=rs .if angle pqr =60degree and angle rpq =70 degree prove that ps is greater than rs and ps is grrayer than pr​

Answer 1

Given :-

• in a triangle PQR ,side PQ is produced to S so that QS = QR.
• ∠PQR = 60° .
• ∠RPQ = 70° .

To Prove :-

• PS > RS .
• PS > PR.

Solution :-

From image , we get,

• ∠PRQ = 180° - (∠PQR + ∠RPQ) = 180° - (60° + 70°) = 180° - 130° = 50° . {Angle sum Property.}
• ∠SQR = 180° - 60° = 120° . {Linear Pair.}
• ∠QSR = ∠QRS = 30° . {QS = QR, and ∠SQR = 120°.}

Now, in ∆PSR, we have,

→ ∠PSR = 30° { = ∠QSR .}

→ ∠SPR = 70° { = ∠RPQ. }

→ ∠PRS = 80° { ∠PRQ + ∠QRS. }

we get,

→ ∠PRS > ∠SPR. {80° > 70°.}

we know,

• The side opposite to the larger angle is longer in length .

Therefore,

→ PS > RS. (Proved.)

Also,

→ ∠PRS > ∠PSR. {80° > 30°.}

Hence,

→ PS > PR. (Proved.)

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Answer 2

Answer:

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