Question

Answer this question!!...​

Answer 1

Step-by-step explanation:

▪️In ∆ QRS, the side SR is produced to T

▪️∴ ∠QRT = ∠RQS + ∠RSQ

▪️[Exterior angle property of a triangle]

▪️But ∠RQS = 28° and ∠QRT = 65°

▪️So, 28° + ∠RSQ = 65°

▪️⇒ ∠RSQ = 65° – 28° = 37°

▪️Since, PQ || SR and QS is a transversal.

▪️∴ ∠PQS = ∠RSQ = 37°

▪️[Alternate interior angles]

▪️⇒ x = 37°

▪️Again, PQ ⊥ PS ⇒ AP = 90°

▪️Now, in ∆PQS,

▪️we have ∠P + ∠PQS + ∠PSQ = 180°

▪️[Angle sum property of a triangle]

▪️⇒ 90° + 37° + y = 180°

▪️⇒ y = 180° – 90° – 37° = 53°

▪️Thus, x = 37° and y = 53°

Hopes it help you✌️✌️

Answer 2

In ∆PQS,

<P + <PQS + <PSQ = 180° (Angle sum property of a ∆)

90° + x + y = 180°

x + y = 90° -(i)

In ∆SQR,

<SQR + <QSR = <QRT (Exterior sum property of a ∆)

28° + <QSR = 65°

<QSR = 37°

<PQS = <QSR (Alternate interior angles as PQ ll SR)

x = 37°

From eq.(i),

37° + y = 90°

y = 53°

Hope it helps!!

Thank you ✌️