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If Two Angles of a Triangle Are Unequal Then the Side Opposite to the Greater Angle Is Greater Than the Side Opposite to Smaller Angle

In the given ...

Question

In the given figure, point S is any point on side QR of $∆$ PQR Prove that : PQ + QR + RP > 2PS

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Solution

In △PQS
PQ + QS > PS ...(1)
(Sum of two sides of a traingle is greater than the third side)
In △PRS
RP + RS > PS ...(2)
(Sum of two sides of a traingle is greater than the third side)
Adding (1) and (2), we get
PQ + QS + RP + RS > PS + PS
⇒ PQ + QR + RP > 2PS

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Q. In the following figure, S is any point on the side QR of Δ PQR. Prove that PQ + QR + RP > 2 PS.

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S is any point on side QR of a

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If Two Angles of a Triangle Are Unequal Then the Side Opposite to the Greater Angle Is Greater Than the Side Opposite to Smaller Angle

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