Lost in the circle

The angle $\angle RQO = \angle ORQ$ , and $\angle QOR=2 \angle QPR$ . Thus $2\angle ORQ+2\angle QPR=180^\circ$ , hence $\angle QPR + \angle ORQ=90^\circ$ .

∠QPR = ∠QP'R = ∠ORP' ∠ORQ + ∠ORP' = 90° Hence ∠ORQ + ∠QPR = 90°

The angle subtended by an arc at the centre is twice the angle subtended by it anywhere else in the circle.

Thus, ∠QOR = 2∠QPR. As ∠QPR= x, therefore ∠QOR= 2x.

Now consider ∆QOR.

OR and OQ are both radii of the circle. Thus, ∆QOR is an isosceles triangle. Angles opposite to equal sides of a triangle are always equal.

Thus ∠ORQ = ∠RQO

But as ∠ORQ= y,

∠RQO = ∠ORQ = y.

Now, using the Angle Sum property of a triangle,

∠QOR + ∠ORQ + ∠RQO = 180 degrees

Substituting in ∠QOR=2x, ∠RQO = y, ∠ORQ = y, we get

2x + y + y = 180°

2x + 2y = 180°

x + y = 90°

Quad Erat Demonstratum.

Without loss of generality, slide P clockwise around the circle, until PR overlaps with OR. Now $\triangle$ PQR is a right triangle, with $\angle$ PQR = 90 $^{\circ}$ . Hence $\angle$ QPR + $\angle$ ORQ = $\angle$ QPR + $\angle$ PRQ = 90 $^{\circ}$

if QOR = X then QPR = 1/2 X, and ORQ = (180 - X)/2 = 90 - (1/2)X Adding gives us:

1/2 X + 90 - 1/2 X = 90

90.(angle ORQ = y & QPR = x therefore angle QOR= 2x because angle subtended by a chord at the center is double the angle subtended by it at any other point on the circle. Now in triangle OQR, angle ORQ =angle OQR = y because OQ and OR are the radius. so 2x +2y = 180. so x + y = 90. and angle ORQ + angle QPR = x + y =90 degree.!!

2(ORQ)+(QOR)=180 (QOR)=180-2(ORQ) * (QPR)=0.5(QOR) (QOR)=2(QPR) * 180-2(ORQ)=2(QPR) (QPR) +(ORQ)= 180\2=90

angle QPR=x angle QOR=2x(central angle) angle OQR=angle ORQ=y thus 2x+2y=180 therefore x+y=90(dividing by 2)

Using radius of circle, triangles OPQ, OQR and OPR are isosceles. Hence, ∠ORP =∠OPR (Let it = A). so, ∠ROP = 180 - 2A (=∠QOP + ∠ROQ).

In triangle OPQ, ∠QOP = 180 - 2(x + A).

In triangle OQR, ∠ROQ = 180 - 2(y)

We said ∠ROP =(∠QOP + ∠ROQ)= 180 - 2A, so......substitute equations for angles

(180 - 2(x+A)) + (180 - 2(y)) = 180 - 2A

Rearranging gives x+y = 90. How I worked it out :p (I'm sure there's an easier way)

Incomplete set of given data. Without constraints, points P, Q, and R can be placed anywhere on the circle, thus, varying the proportions and values of the angles.

$\angle OQR = \angle ORQ = y \implies \angle QOR = 180-2y = 2x \implies x = \frac{180-2y}{2} = 90-y$

$\implies x+y = 90-y+y = \boxed{90^\circ}$

Atention! They've marked in pink something wrong (PRQ instead ORQ). That can trick you...

Since we have choices, angle Q is greater than 90 degrees and a triangle is equal to 180 degrees, so it must be the sum of the 2 other angles is less than or equal to 90 degrees..

Since both triangles QRP and QRO have the same base, the angle subtended by an arc in the centre is twice the angle subtended by this arc anywhere on the circumference. Hence, $\angle ROQ = 2\times \angle RPQ$ . Since $RO\ = QO$ , $\angle QRO$ can be written as $\dfrac{180 - 2\times\angle RPQ}{2}$ If we then add $\angle RPQ$ to the result: $90 - \angle RPQ + \angle RPQ\ = \boxed{90}$

As point Q is moved towards R angle ORQ approaches a 90 degree angle and RPQ approaches 0 and as point Q moves towards point P the opposite happens. Given the angle choices 90 was the obvious solution.

Angle OQR=Angle ORQ[ OQ=OR, radii of same circle ] But angleORQ=y, so angle OQR=y. Now angle QOR=180-2y and angleQPR=1/2QOR. So we get 180-2y/2=x. On solving this equation, we get 90-y=x. So, x+y=90

Angle ROQ=2x y=180-2x/2 x+y=x+180-2x/2=2x+180-2x/2 =180/2=90

QOR=2QPR & OQR=ORQ, from triangle, QOR, QOR+OQR+ORQ=180, OR, 2QPR+ORQ+ORQ=180, OR,2QPR+2ORQ=180, OR,x+y=90

The angle at the center that is subtended by an arc is twice any angle subtended by the same arc anywhere else in the circle. Since $\angle QOR$ is subtended by $\widehat{QR}$ , and is at the center of the circle, that means that $\angle QOR = 2\angle QPR$ since $\angle QPR$ is also subtended by $\widehat{QR}$ . Since $\overline{OQ}$ and $\overline{OR}$ are both radii of the same circle, they are congruent. Because of this, the triangle that they create with $\overline{QR}$ is an isosceles triangle. Since the base angles of an isosceles triangle are always equal, $\angle OQR=\angle ORQ$ . We now have the 3 angles in Isosceles triangle $QOR$ : $\angle OQR,\angle ORQ$ , and $\angle QOR$ .

Finding x: We can use the substitution property to replace $\angle QPR$ with $x$ , so we now get the statement $\angle QOR = 2x$ .

So now we have isosceles triangle $QOR$ . If we were to add the three angles in it, we'd get $2x$ (from the vertex angle, $\angle QOR$ ) $+y+y$ (Because there are two base angles, which both equal y since they are equal). Because the sums of the interior angles of a triangle is always $180 ^ \circ$ , we can create the simplified equation $2x+2y+180$ . All we have to do now is divide both sides by 2 to get the equation $x+y=90$ . So now it is clear to see that 90 is the answer!

Since we know the angles opposite to the equal sides of a triangle are equal so, angORQ=angOQR. Again angQOR=2angQPR. Now, Q+O+R=180. or, 2y+2x=180 Therefore, x+y=90 degrees.

QPR=x OQR=2x ORQ=(180-2x)/2=90-x=y

x+y=x+(90-x)=90

x=1/2 O (because any angle suspended by an arc on any point of circle is half to that of angle suspended by that arc in the centre ). Q=R now, O+Q+R=180 therefore, y= 180-O/2

therefore, x+y= 90-(O/2)+(O/2) thus, x+y= 90

2x + 2 y = 180 so.... x + y = 90