NCERT Solution: Lines and Angles
In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Answer
Given,
PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°
A/q,
x +∠SQR = ∠QRT (Alternate angles as QR is transveersal.)
⇒ x + 28° = 65°
⇒ x = 37°
also,
∠QSR = x
⇒ ∠QSR = 37°
also,
∠QRS +∠QRT = 180° (Linea pair)
⇒ ∠QRS + 65° = 180°
⇒ ∠QRS = 115°
Now,
∠P + ∠Q+ ∠R +∠S = 360° (Sum of the angles in a quadrilateral.)
⇒ 90° + 65° + 115° + ∠S = 360°
⇒ 270° + y + ∠QSR = 360°
⇒ 270° + y + 37° = 360°
⇒ 307° + y = 360°
⇒ y = 53°
In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2∠QPR.
AnswerGiven,
Bisectors of ∠PQR and ∠PRS meet at point T.
To prove,
∠QTR = 1/2∠QPR.
Proof,
∠TRS = ∠TQR +∠QTR (Exterior angle of a triangle equals to the sum of the two interior angles.)
⇒ ∠QTR = ∠TRS - ∠TQR --- (i)
also,
∠SRP = ∠QPR + ∠PQR
⇒ 2∠TRS = ∠QPR + 2∠TQR
⇒ ∠QPR = 2∠TRS - 2∠TQR
⇒ 1/2∠QPR = ∠TRS - ∠TQR --- (ii)
Equating (i) and (ii)
∠QTR - ∠TQR = 1/2∠QPR
Hence proved.