Question

If $PR$ is a diameter of the circle and the angles $RPS,QPR,QRPandPRS$ forms an $AP$ with $RPS={15}^{\circ }$, then the value of theangle $QRP$ is

• A. ${55}^{\circ }$
• B. ${75}^{\circ }$
• C. ${60}^{\circ }$
• D. ${35}^{\circ }$

Answer 1

The correct option is A ${55}^{\circ }$

Given angle RPS, QPR, QRP and PRS forms an AP.

Also, RPS = 15

Clearly, 2QPR = RPS + QRP [According to arithmetic progression]

or, 2QPR = QRP + 15 .... (1)

In triangle PRS, we have

Angle PSR = 90 [Angle in a semi-circle is a right angle]

So, by angle sum property of triangle, we have

PSR + PRS + RPS = 180

or, 90 + PRS + 15 = 180

or, PRS = 75

Again, 2QRP = PRS + QPR [According to arithmetic progression]

or, 2QRP = QPR + 75

or, 4 QRP = 2QPR + 150 ... (2)

Putting (1) in (2), we get

4QRP = QRP + 15 + 150

or, 3QRP = 165

or, QRP = 55

Hence, option (A) is the correct answer.