Question

In fig. PT and PQ are tangents to the circle from an external point....
Full questions is in the attachment plzz help.

Answer 1

Answer:

∠TRQ = 55°

Step-by-step explanation:

From figure:

⇒ ∠OTP = ∠OQP = 90° {Radius and tangent are perpendicular at their point of contact}

In Quadrilateral OQPT,

⇒ ∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360°

⇒ ∠QOT + 90° + 90° + 70° = 360°

⇒ ∠QOT + 250° = 360°

⇒ ∠QOT = 110°

Now,

⇒ ∠TRQ = (1/2) * ∠QOT

              = (1/2) * 110°

              = 55°

Hope it helps!

Answer 2

Now, In quadrilateral OQPT

∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360° [Angle sum property of a quadrilateral] ∠QOT + 90° + 90° + 70° = 360°

250° + ∠QOT = 360°

∠QOT = 110°

We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.

∠TRQ = 12 and ∠QOT=55°