Question

solve this guys...........★​

Answer 1

Answer:

XYP is a straight line.

∴ ∠XYZ + ∠ZYQ + ∠QYP = 180°

⇒ 64° + ∠ZYQ + ∠QYP = 180°

[∵ ∠XYZ = 64° (given)]

⇒ 64° + 2∠QYP = 180°

[YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ]

⇒ 2∠QYP = 180° – 64° = 116°

⇒ ∠QYP = 116°/2 = 58°

∴ Reflex ∠QYP = 360° – 58° = 302°

Since ∠XYQ = ∠XYZ + ∠ZYQ

⇒ ∠XYQ = 64° + ∠QYP [∵∠XYZ = 64°(Given) and ∠ZYQ = ∠QYP]

⇒ ∠XYQ = 64° + 58° = 122° [∠QYP = 58°]

Thus, ∠XYQ = 122° and reflex ∠QYP = 302°.

This is your answer hope it helps you please mark it as brainlyest ....

Answer 2

Answer:

Exterior Angle of a triangle:

If a side of a triangle is produced then the exterior angle so formed is equal to the sum of the two interior opposite angles.

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Solution:

Given,

Bisectors of ∠PQR & ∠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)

∠SRP = ∠QPR + ∠PQR

⇒ 2∠TRS = ∠QPR + 2∠TQR

[ TR is a bisector of ∠SRP & QT is a bisector of ∠PQR]

⇒∠QPR= 2∠TRS – 2∠TQR

⇒∠QPR= 2(∠TRS – ∠TQR)

⇒ 1/2∠QPR =  ∠TRS – ∠TQR — (ii)

Equating (i) and (ii)

∠QTR= 1/2∠QPR

Hence proved.

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