Question

In ΔPQR, ∠Q = 70°, ∠QPR = 50° and PR = RS. Prove that PQ < PS.

Answer 1

Given:

In ΔPQR, ∠Q = 70°, ∠QPR = 50°

PR = RS (point S lies on the extension of QR such that PR = RS)

To Prove: PQ < PS

In ΔPQR, ∠Q = 70°, ∠QPR = 50° and PR = RS.

∠P + ∠Q + ∠R = 180°

50° + 70° + ∠R = 180°

∠R = 60°   ...[180° – (∠P + ∠Q) = 180° – (50° + 70°)]

So, ∠PRS = 180° – ∠R (60°)   ...(Linear pair)

∠PRS = 120°   ...[180° – 60°]

Then, ∠PRS + ∠RPS + ∠PSR = 180°   ...[Defined through angle sum property]

∠RPS = ∠PSR   ...[Property of isosceles triangle]

Then, 120° + ∠RPS + ∠PSR = 180°

∠RPS + ∠PSR = 60°   ...(∠RPS = ∠PSR)

2 × ∠PSR = 60°

∠PSR = 30°

Therefore, ∠QPS = ∠QPR + ∠RPS = 50° + 30° = 80°

Angle Q = 70°

Angle S = 30°

Therefore, ∠Q < ∠S

So, PQ < PS

Hence, proved.