In the given diagram, $O$ is the center of the circle. We are given that $\angle SQR = 60^{\circ}$, $\angle SPR = y$, and $\angle SOR = 3x$. We need to find the value of $x + y$.

1. Problem Description

In the given diagram, $O$ is the center of the circle. We are given that $\angle SQR = 60^{\circ}$, $\angle SPR = y$, and $\angle SOR = 3x$. We need to find the value of $x + y$.

2. Solution Steps

Since $O$ is the center of the circle, $\angle SOR$ is the central angle subtended by the arc $SR$. $\angle SPR$ is an inscribed angle subtended by the same arc $SR$.

We know that the inscribed angle is half of the central angle. Therefore, we have:

$y = \frac{1}{2} \angle SOR$

$y = \frac{1}{2} (3x)$

$y = \frac{3}{2}x$

Also, we know that the angle at the center is twice the angle at the circumference subtended by the same arc. Therefore:

$\angle POR = 2\angle PSR$

Since $OS=OR$ (radii of the same circle), $\triangle SOR$ is an isosceles triangle. Thus $\angle OSR = \angle ORS$.

Also, we have $\angle SOR + \angle OSR + \angle ORS = 180^{\circ}$.

Since $\angle SOR = 3x$, we have $3x + \angle OSR + \angle ORS = 180^{\circ}$.

Since $\angle OSR = \angle ORS$, we have $3x + 2\angle ORS = 180^{\circ}$.

$\angle ORS = \frac{180 - 3x}{2} = 90 - \frac{3}{2}x$.

We are given that $\angle SQR = 60^{\circ}$. Also, $\angle SQR$ and $\angle SPR$ subtend the same arc SR. Thus $y = \angle SPR$. Since $y = \frac{1}{2} (3x)$, $2y = 3x$.

Now consider the quadrilateral $PSQR$.

The angles $\angle PSR$ and $\angle PQR$ are subtended by the same chord $PR$.

Also $\angle PSR + \angle PRQ = 180^{\circ} - (\angle QPS + \angle SQR)$

We are given $\angle SQR = 60^{\circ}$. Also $OS = OR$, which implies triangle $SOR$ is an isosceles triangle. Therefore $\angle OSR = \angle ORS = (180^{\circ}-3x)/2 = 90-1.5x$. Similarly $OP=OQ$, which means $\angle OPQ = \angle OQP$. We are given $\angle SQR = 60^{\circ}$.

Consider the angle $\angle SOR=3x$. The inscribed angle that subtends the same arc is $y$. $2y = 3x$, so $y = \frac{3}{2} x$.

In quadrilateral $PSQR$, $\angle SPR = y$ and $\angle SQR = 60^{\circ}$. The sum of the angles of quadrilateral is

3

6

0. We can consider the $\angle SOR$. $\angle SOR = 2 \angle SQR$.

$\angle SPR = \frac{\angle SOR}{2} = \frac{3x}{2} = y$.

Since the total angles in the quadrilateral $PSQR$ add up to $360^{\circ}$,

$\angle SQR = 60^{\circ}$

$\angle SPR = y$

$\angle PSR = a$

$\angle PQR = b$

$y + 60 + a + b = 360$

$a + b = 300 - y$.

Consider triangle $OSR$. $OS=OR$ and $\angle SOR = 3x$. Therefore $\angle OSR = \angle ORS$.

$\angle OSR = \angle ORS = (180-3x)/2 = 90 - 1.5 x$.

We are given $\angle SQR = 60$.

We also have inscribed angle $y$ corresponds to $3x$ at center.

$\implies 2y = 3x$

Since the arc SR subtends angle $SQR = 60$ at $Q$, we have $\angle SOR= 2\angle SQR=120^{\circ}$.

Then $3x=120^{\circ}$, $x=40^{\circ}$. Since $2y=3x$, we have $2y=3(40)=120^{\circ}$. Therefore $y=60^{\circ}$.

Therefore $x+y = 40^{\circ}+60^{\circ}=100^{\circ}$.

3. Final Answer

100°