Fewpal
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In the figure, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ

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In the given figure, ∠SPR = 135° and 

∠PQT = 110°. ∠PQT + ∠PQR = 180° [Linear pair axiom] 

⇒ 110° + ∠PQR = 180°

⇒ ∠PQR = 180° – 110° = 70° 

Also, ∠SPR + ∠QPR = 180° [Linear pair axiom] 

⇒ 135° + ∠QPR = 180° 

⇒ ∠QPS = 180° – 135° = 45° 

Now, in the triangle PQR 

∠PQR + ∠PRQ + ∠QPR = 180° 

[Angle sum property of a triangle] 

⇒ 70° + ∠PRQ + 45° = 180° 

⇒ ∠PRQ + 115° = 180° 

⇒ ∠PRQ = 180° – 115° = 65° 

Hence, ∠ PRQ = 65° Ans.

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