Find The PRO Angle

First of all, Join OQ. Now, BO = RQ = OQ (radius of same circle). So, $\angle$ QRO = $\angle$ QOR. Let them be x.

Applying exterior angle property, $\angle$ PQO = $\angle$ QRO + $\angle$ QOR = 2x. And as OQ = OP, $\angle$ PQO = $\angle$ QPO = 2x. And using angle sum property, $\angle$ POQ = 180 - 4x.

Now AB is a Straight line. So, $\angle$ AOQ + $\angle$ QOP + $\angle$ BOP = 180. And we conclude that $\angle$ BOP = 3x. But $\angle$ BOP = 60 and this enables us to find the value of x = 20.

So, $\angle$ PRO = $\boxed{20}$ .

As other answers have explained, $\bigtriangleup QRO$ is isosceles with $QR = QO$ . Let $x$ denote $\angle QRA$ and $\angle QOA$ (they are congruent).

There is a very useful identity on the angles formed by two secant lines in a circle. Secant angle identity

Applying the identity: $x = \frac12 ( \angle POB - \angle QOA)$

$x = \frac12 (60^\circ - x)$

$\boxed {x = 20^\circ }$

Let the desired angle be 't'. So by sine rule of the triangle poq which is isosceles we get that sin(60-t)=sin2t.So we get that t is 20 or 120 degrees. But t can not be obtuse.so t=20 degrees....

BOP is an equilateral triangle; let angle OPR=y; angle PRO=x; x+y=60 by exterior angle theorem(1); triangle POQ is isosceles and hence angle PQO = y; in triangle QRO 2x+180-y = 180 -> y = 2x; hence 3x=60 by (1) -> x = 20

join OQ. now, RQ=BO (given) also BO=OQ= Radius .: RQ=OQ join OQ=OP=RADIUS NOW APPLYIN ANGLE SUM PROP AND PROPERTIES OF ISOSCELES TRIANGLE ONE CAN SOLVE ANGLE PRO=20

There is an isosceles triangle RQO since QO=BO=RQ.

The angles opposite the equal sides are also equal, so angle QRA = angle QOA = arc AQ (central angle), let them be x.

To find the angle (x) formed by two secant lines that intersect in the exterior of a circle is... one half of the difference of the measure of the intercepted arcs.

x = 1/2 (60 - x)

x = 20

Simply join the "point O" with "point Q", then it will make a isoceles triangle and the angles <ORQ=<ROQ. Then name these angle with any given name let $\theta$ . Then you know that the angle <QOR will be 120- $\theta$ . Then simply find the chord length QR since by then you would know the angle inscribed by the chord. After that you would know the length of segment RP wich will RQ+QR. Then by using sine formula, one can easily solve for the given angle $\theta$ .

well it was a question with lack of information..even though i solved it s=r€ is the answer compare and get it ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, let oA=oB=rQ=m=RA,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,s=r€ in OPB ......................WE GET..........>S=m60 ,,,,,,,,,,,now in RAQ,,,,,,,,,,,,,,,s=r€,,,,,,,,,,,,,,,,,,,,>>>>.s=3m€,,,,,,,,,,,,, ,,,,,,,,,,,,,,,now comparing s.........................3m€=m60.

>>>>>>>>>>>>>.€=20