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q.6 ∠TRS is an exterior angle of △TQR
∠TRS=∠TQR+∠QTR …..(i) (Since, the exterior angle is equal to the sum of the two interior opposite angles)
∠PRS is an exterior angle of △PQR
∠PRS=∠PQR+∠QPR ……(ii) (since, the exterior angle is equal to the sum of the two interior opposite angles)
⇒2∠TRS=2∠TQR+∠QPR (QT is the bisector of ∠PQRandRT is the bisector of ∠PRS⇒∠TRS
⇒2(∠TRS−∠TQR)=∠QPR ……(iii)
From (i), ∠TRS−∠TQR=∠QTR ……(iv)
From (iii) and (iv) , we obtain
2∠QTR=∠QPR ⇒∠QTR=12∠QPR
∠TRS=∠TQR+∠QTR …..(i) (Since, the exterior angle is equal to the sum of the two interior opposite angles)
∠PRS is an exterior angle of △PQR
∠PRS=∠PQR+∠QPR ……(ii) (since, the exterior angle is equal to the sum of the two interior opposite angles)
⇒2∠TRS=2∠TQR+∠QPR (QT is the bisector of ∠PQRandRT is the bisector of ∠PRS⇒∠TRS
⇒2(∠TRS−∠TQR)=∠QPR ……(iii)
From (i), ∠TRS−∠TQR=∠QTR ……(iv)
From (iii) and (iv) , we obtain
2∠QTR=∠QPR ⇒∠QTR=12∠QPR
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q.5 ∠QRT=∠RQS+∠QSR (exterior angle is equal to sum of the two opposite interior angles)
⇒65∘=28∘+∠QSR ⇒∠QSR=65∘−28∘=37∘
PQ⊥SP (given)
∠QPS+∠PSR=180∘ (the sum of consecutive interior angles on the same side of the transversal in 180∘ )
⇒90∘+∠PSR=180∘ ⇒∠PSR=180∘−90∘=90∘ ⇒∠PSQ+∠QSR=90∘ ⇒y+37∘=90∘ ⇒Y=90∘−37∘=53∘
In △PSQ,∠PSQ+∠QSP+∠QPS=180∘ (as sum of interior angles of a triangle is 180∘.
⇒x+y+90∘=180∘ x+53∘+90∘=180∘ x+143∘=180∘ x=180∘−143∘=37∘
rqs kaise aya
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