Question

In figure 3.91, line PR touches the circle at point Q. Answer the following questions with the help of the figure.
(1) What is the sum of ∠TAQ and ∠TSQ ?
(2) Find the angles which are congruent to ∠AQP.
(3) Which angles are congruent to ∠QTS ?
(4) ∠TAS = 65°, find the measure of ∠TQS and arcTS
(5) If ∠AQP=42°and ∠SQR=58° find measure of ∠ATS.

Answer 1

Jack in the figure out what you want me to

Answer 2

Answer:

(1) ∠TAQ + ∠TSQ = 180°

(2) ∠AQP is congruent to ∠ASQ and ∠ATQ

(3) ∠QTS is congruent ∠QAS and ∠SQR

(4) ∠TQS = 65°

(5) ∠ATS = 100°

Step-by-step explanation:

(1) Since the vertices of the quadrilateral ATSQ lie on the circle, therefore the quadrilateral ATSQ is a cyclic quadrilateral.

We know that in cyclic quadrilateral the sum of the opposite angles is 180°

∠TAQ and ∠TSQ are opposite angles of cyclic quadrilateral ATSQ

Therefore, ∠TAQ + ∠TSQ = 180°

(2) ∠AQP is the angle made between the chord AQ and tangent PR. By theorem this angle will be equal to the angle made by chord AQ on the circle. These angles are ∠ASQ and ∠ATQ

Therefore, ∠AQP = ∠ASQ = ∠ATQ

i.e. ∠AQP is congruent to ∠ASQ and ∠ATQ

(3) ∠QTS is the angle made by chord QS on the circle. This angle will be equal to the another angle made by QS on the same side of chord in the circle. This angle is ∠QAS. Also the angle ∠QTS will be equal to the angle between chord QS and the tangent PR which is ∠SQR

Therefore, ∠QTS = ∠QAS = ∠SQR

i.e. ∠QTS is congruent ∠QAS and ∠SQR

(4) Given ∠TAS = 65°

∠TAS is the agle made by chord TS which will be equal to another angle ∠TQS made by the chord TS on the the same side of the circle

Thus, ∠TAS = ∠TQS = 65°

Now the angle made by the chord TS on the centre of the circle will be twice the angle made by the chord on the circle (by theorem)

The angle made by chord TS on the centre = 2 × 65° = 130°

Thus, the length of arc TS

$=\frac{130^\circ\times \pi}{180^\circ}\times R$  (Where R is radius of the circle)

$=\frac{13\pi R}{18}$

(5) Given

∠AQP=42°and ∠SQR=58°

Since PQ is a straight line

Therefore by linear pair axiom

∠AQP + ∠AQR = 180°

or, ∠AQP + ∠AQS + ∠SQR = 180°

or, 42° + ∠AQS + 58° = 180°

or, 100° + ∠AQS = 180°

or, ∠AQS = 80°

Since the quadrilateral ATSQ is a cyclic quadrilateral

Therefore, ∠AQS + ∠ATS = 180°

or, ∠ATS = 180° - ∠AQS

               = 180° - 80°

               = 100°

Hope this helps.