In the given figure O is the centre of the circle. If ∠PAB = 35°, then find ∠ARS.

Given: 

O is the center of the circle.

​$$\angle PAB = 35^\circ$$ 

Concept Used: 

Angles made on the circle by the same arc is equal.

Straight angle = $$180^\circ$$ 

Sum of all the angle in a triangle = $$180^\circ$$  

Angle made by diameter on a circle = $$90^\circ$$​​

In $$\triangle APB$$ 

$$\angle APB = 90^\circ$$( angle by diameter)

Now, 

​$$\angle PAB + \angle ABP + \angle APB = 180^\circ \\ \angle ABP + 35 + 90 = 180 \\ \angle ABP = 180 - 125 \\ \angle ABP = 55^\circ$$ 

since, $$\angle ABP = \angle ARP = 55^\circ$$ ( angle by same arc AP)

SRP is a straight angle,

​$$\angle SRA + \angle PRA = 180^\circ \\ \angle SRA = 180^\circ - 55^\circ \\ \angle SRA = 125^\circ$$ 

Thus, $$\bf \angle ARS = 125^\circ$$​​