Question

15. In triangle PQR, angle Q = 70°, angle QPR = 50° and PR=RS.
Prove that PQ <PS.​

Answer 1

Answer:

Answer in the given attachment.

Please mark my answer as the brainliest.

Answer 2

Answer:

PQ > PS

∆P + ∆Q + ∆R = 180°

= 50° + 70° + ∆R = 180°

= ∆R = 60° [180° - (∆P + ∆Q) = 180° - (50° + 70°)]

So ∆PRS = 180° - ∆R(60°)

= ∆PRS = 120° [180° - 60°]

Then,

∆ PRS + ∆RPS + ∆PSR = 180° (defined through angle sum property)

∆RPS = ∆PSR [property of isosceles ∆]

Then,

∆RPS + ∆PSR + 120° = 180°

∆PSR + ∆PRS = 60°( ∆RPS < ∆PSR)

2 × ∆PSR = 60°

∆PSR = 30°

Therefore

∆QPS = 50° + 30° = 80°

angle Q = 70°

angle S = 30°

Therefore

angle Q < angle S

So,

PQ < PS

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