Question

answer it fast please​

Answer 1

Step-by-step explanation:

Here, O is the centre of circle.

PQ and PT are tangents to the circle from a point P

R is any point on the circle. RT and RQ are joined.

∠TPQ=70⁰

Now,

Join TO and QO

∠TOQ=180⁰ −70 =110⁰

Here, OQ and OT are perpendicular on QP and TP.

∠TOQ is on the centre and ∠TRQ is on the rest part.

∠TRQ=1/2∠TOQ=1/2(110⁰)=55⁰

Therefore , ∠TRQ=55 degrees.

Answer 2

$angle \: rtp \: and \: angle \: rqt = {90}^{o} \\ angle \: rtp + \: angle \: rqt + \: angle \: tpq + \: angle \: trq = {360}^{o} \\ {90}^{o} + {90}^{o} + {70}^{o} + angle \: trq = {360}^{o} \\ angle \: trq = {360}^{o} - {250}^{o} \\ angle \: trq = {110}^{o}$