Circle Theorems (B) Worksheet | Fun and Engaging Geometry PDF ... - Free Printable

Here are the step-by-step solutions for each problem on the worksheet. I have used standard circle theorems to find the missing angles.

1)
* Angle a: In a cyclic quadrilateral (a four-sided shape inside a circle), opposite angles add up to $180^\circ$. Angle $a$ is opposite the $106^\circ$ angle.
$180^\circ - 106^\circ = 74^\circ$.
* Angle b: Angle $b$ is opposite the $93^\circ$ angle.
$180^\circ - 93^\circ = 87^\circ$.

2)
* Angle c: This is a cyclic quadrilateral. Angle $c$ is opposite angle $d$. We need to find angle $d$ first or use the exterior angle property. The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. So, angle $c$ is equal to the exterior angle at the bottom right? No, let's look closer.
Let's find the interior angle next to the $102^\circ$ exterior angle first. Angles on a straight line add to $180^\circ$.
Interior angle $= 180^\circ - 102^\circ = 78^\circ$.
Now, opposite angles in a cyclic quad sum to $180^\circ$.
Angle $c + 78^\circ = 180^\circ \rightarrow c = 102^\circ$. (Alternatively, the exterior angle equals the interior opposite angle, so $c = 102^\circ$).
* Angle d: Angle $d$ is opposite the $69^\circ$ angle.
$180^\circ - 69^\circ = 111^\circ$.

3)
* Angle e: The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. The interior opposite to angle $e$ is $88^\circ$. So, $e = 88^\circ$.
* Angle f: First, find the interior angle adjacent to the $91^\circ$ exterior angle. $180^\circ - 91^\circ = 89^\circ$.
Opposite angles sum to $180^\circ$. So, $f + 89^\circ = 180^\circ$.
$f = 180^\circ - 89^\circ = 91^\circ$. (Or simply, exterior angle $91^\circ$ = interior opposite angle $f$).

4)
* Angle g: The triangle containing angle $g$ has one side as the diameter. The angle subtended by a diameter at the circumference is always $90^\circ$ (Angle in a semicircle). So, the angle at the bottom left vertex is $90^\circ$.
Wait, looking at diagram 4, the line with the dot is the diameter. The triangle formed by the diameter and point $g$ is a right-angled triangle.
Actually, let's look at the triangle with vertices: center, bottom-left point, top-right point ($h$).
Let's use the property that angles in the same segment are equal? No.
Let's look at the triangle containing angle $g$. It shares the diameter. The angle opposite the diameter is $90^\circ$. So the angle at vertex $h$'s neighbor? No.
Let's identify the right angle. The angle subtended by the diameter at the circumference is $90^\circ$.
In the triangle with vertices at the ends of the diameter and the top vertex (angle $g$), the angle at the third vertex isn't given.
Let's look at the triangle with vertices: Left point, Right point ($h$), and Bottom point ($37^\circ$). The side connecting Left and Right is the diameter. Therefore, the angle at the bottom vertex is NOT necessarily $90^\circ$ unless the bottom vertex is on the circle and the side is the diameter. Yes, the line goes through the center. So the angle at the vertex on the circle opposite the diameter is $90^\circ$.
Which triangle has the diameter as a side? The triangle with vertices: Top-Left ($g$), Bottom-Left ($37^\circ$), and Top-Right ($h$). The side connecting Bottom-Left and Top-Right is the diameter.
Therefore, the angle at the remaining vertex on the circumference (Top-Left, angle $g$) is $90^\circ$? No, the angle *subtended* by the diameter is $90^\circ$.
The angle at vertex $g$ is subtended by the chord connecting the other two points. That's not the diameter.
The diameter connects the bottom-left point and the top-right point.
Therefore, the angle at the top-left point ($g$) is NOT the one subtended by the diameter.
The angle at the bottom-right point (let's call it $y$) would be $90^\circ$ if we drew a line there.
Let's look at the triangle formed by the diameter and the point with angle $37^\circ$. The vertices are Center, Bottom-Left, Top-Right. This is an isosceles triangle.
Actually, simpler approach:
Angle $h$ and angle $37^\circ$ are angles in the same segment? No.
Let's look at the triangle with the diameter as the hypotenuse. The vertices are the bottom-left point, the top-right point, and... wait.
The line segment passing through the center connects the bottom-left vertex and the top-right vertex.
Therefore, the angle at the top-left vertex ($g$) and the angle at the bottom-right vertex are both $90^\circ$ because they are angles in a semicircle.
So, $g = 90^\circ$.
Now for $h$: Consider the triangle formed by the top-left, top-right, and bottom-left vertices.
We know angle $g = 90^\circ$.
We know the angle at the bottom-left is split? No, the $37^\circ$ is the whole angle at that vertex for the triangle involving the diameter?
Let's re-read diagram 4.
There is a triangle inscribed in the semicircle. Vertices: Bottom-Left, Top-Left ($g$), Top-Right ($h$).
The side Bottom-Left to Top-Right is the diameter.
Therefore, angle $g = 90^\circ$.
Now consider the triangle Bottom-Left, Top-Right, and... wait, where is $37^\circ$?
The $37^\circ$ is at the bottom-left vertex. It is part of the triangle with vertices Bottom-Left, Top-Left, Top-Right?
If so, the sum of angles in that triangle is $180^\circ$.
Angle at Bottom-Left $= 37^\circ$.
Angle at Top-Left ($g$) $= 90^\circ$.
Angle at Top-Right ($h$) $= 180^\circ - 90^\circ - 37^\circ = 53^\circ$.
So, $h = 53^\circ$.

5)
* Angle i: The angle at the center ($j$) and the angle at the circumference ($i$) subtend the same arc? No.
Let's look at the triangle formed by the center and two points on the circumference. It is an isosceles triangle because two sides are radii.
However, there is a simpler theorem: The angle at the center is double the angle at the circumference subtended by the same arc.
Angle $j$ is at the center. Angle $70^\circ$ is at the circumference. They subtend the same arc (the one on the left).
So, $j = 2 \times 70^\circ = 140^\circ$.
Now, look at angle $i$. It is an angle in a triangle with the center?
Actually, let's look at the "bowtie" or "butterfly" shape. Angles subtended by the same arc at the circumference are equal.
The angle labeled $70^\circ$ subtends the arc between the top-left and bottom-left points.
Angle $i$ subtends the arc between the top-left and bottom-left points? No.
Let's trace the lines.
Angle $70^\circ$ is formed by chords from Top-Left to Bottom-Left and Top-Left to Center? No, the vertex is on the circle.
Let's assume the standard configuration:
Angle at circumference $= 70^\circ$. Subtends arc $AB$.
Angle at center $j$ subtends arc $AB$. So $j = 2 \times 70^\circ = 140^\circ$.
Angle $i$ is in a triangle with the center? The triangle is Isosceles (two radii).
The angle at the center facing angle $i$ is vertically opposite to part of $j$? No.
Let's look at the triangle containing $i$. Vertices: Center, Top-Right, Bottom-Right?
Actually, angle $i$ and the $70^\circ$ angle subtend the same arc if they are in the same segment.
Looking at the lines: The chord from Top-Left to Bottom-Right passes through the center? No.
Let's look at the arc subtended by angle $i$. It's the arc from Top-Left to Bottom-Left.
The angle $70^\circ$ also subtends the arc from Top-Right to Bottom-Right?
Let's restart Diagram 5 carefully.
We have a circle with center. Two chords intersecting? No.
We have a triangle with vertices on the circle: Top-Left, Bottom-Left, Top-Right.
And a line from Bottom-Left to Center to Top-Right? That would make it a diameter.
If the line from Bottom-Left to Top-Right is a diameter, then the angle at Top-Left is $90^\circ$. But it's labeled $70^\circ$? No, the $70^\circ$ is at the Top-Left vertex.
If the line is a diameter, the angle in the semicircle is $90^\circ$. The diagram shows a triangle with a vertex at the center.
Okay, Triangle 1: Vertices Center, Top-Left, Bottom-Left. This is isosceles.
Triangle 2: Vertices Center, Top-Right, Bottom-Right?
Let's look at angle $i$. It is at the Top-Right vertex.
The angle $70^\circ$ is at the Top-Left vertex.
They both subtend the same arc at the bottom?
The chords forming the $70^\circ$ angle go to the Bottom-Left and... somewhere else.
The chords forming angle $i$ go to the Bottom-Left and... somewhere else.
Actually, usually in these diagrams, angles in the same segment are equal.
If the $70^\circ$ angle and angle $i$ both stand on the same arc (the bottom arc), then $i = 70^\circ$.
Let's check the lines.
Line from Top-Left to Bottom-Right.
Line from Top-Right to Bottom-Left.
These intersect at the center? If they intersect at the center, they are diameters.
If they are diameters, then the triangle formed by Top-Left, Center, Top-Right is isosceles.
And the angle $70^\circ$ is part of a right triangle?
Let's assume the line passing through the center is a straight line (diameter).
The diagram shows a line from Bottom-Left to Top-Right passing through the center. So that is a diameter.
Therefore, the angle at the Top-Left vertex (subtended by the diameter) should be $90^\circ$. But the label $70^\circ$ is inside a smaller triangle?
Ah, the $70^\circ$ is the angle $\text{Top-Left} - \text{Bottom-Left} - \text{Center}$? No, the vertex is on the circle.
Let's assume the standard "Angle at center is twice angle at circumference".
Angle $j$ is the central angle.
Angle $70^\circ$ is the circumference angle subtending the same arc as angle $j$?
If yes, $j = 140^\circ$.
Then angle $i$: Look at the triangle with the center. It is isosceles.
The angle at the center for that triangle is $180^\circ - 140^\circ = 40^\circ$ (angles on a straight line).
The triangle containing $i$ has vertices: Center, Top-Right, Bottom-Right?
Or is $i$ simply the angle in the same segment as another angle?
Let's look at the arc between Top-Left and Bottom-Left.
Angle at center $= j$.
Angle at circumference $= 70^\circ$? No, the $70^\circ$ vertex is Top-Left. The rays go to Bottom-Left and Bottom-Right?
If the rays go to Bottom-Left and Bottom-Right, and the center is involved...
Let's try this interpretation:
The angle $70^\circ$ subtends the arc at the bottom.
Angle $i$ also subtends the arc at the bottom.
Therefore $i = 70^\circ$ (Angles in the same segment are equal).
Then angle $j$: This is the central angle subtending the same arc.
Therefore $j = 2 \times 70^\circ = 140^\circ$.

6)
* Angle k: The triangle has the diameter as one side (the line with the dot). Therefore, the angle opposite the diameter is $90^\circ$.
The angle at the top vertex is $90^\circ$.
The sum of angles in a triangle is $180^\circ$.
The triangle vertices are: Bottom-Left ($k$), Top-Right ($51^\circ$?), and Top-Left ($90^\circ$).
Wait, the $51^\circ$ is at the Top-Right vertex.
So, $k + 51^\circ + 90^\circ = 180^\circ$.
$k = 180^\circ - 141^\circ = 39^\circ$.
* Angle l: Angle $l$ and angle $k$ subtend the same arc? No.
Angle $l$ is in the same segment as the angle at the top-left?
Let's look at the arc subtended by angle $l$ (Bottom-Right). It subtends the arc from Top-Left to Bottom-Left.
The angle at the Top-Right ($51^\circ$) subtends the arc from Top-Left to Bottom-Left.
Therefore, angles in the same segment are equal.
$l = 51^\circ$.

7)
* Angle m: This is a cyclic quadrilateral.
Opposite angles sum to $180^\circ$.
Angle $m$ is opposite the $67^\circ$ angle.
$m = 180^\circ - 67^\circ = 113^\circ$.
* Angle n: Angle $n$ is opposite the $76^\circ$ angle.
$n = 180^\circ - 76^\circ = 104^\circ$.

8)
* Angle p: The quadrilateral has tick marks on all four sides, meaning it is a rhombus (or square) inscribed in a circle. A rhombus inscribed in a circle must be a square.
Therefore, all corner angles are $90^\circ$.
However, let's verify with theorems.
The diagonal passes through the center, so it is a diameter.
The angle subtended by the diameter at the circumference is $90^\circ$.
So the angle at vertex $p$ (which is part of the triangle formed by the diameter) is $90^\circ$?
Vertex $p$ is a corner of the quadrilateral.
The diagonal splits the angle.
Since it's a square (equal chords imply equal arcs, $360/4 = 90^\circ$ per arc), the angle at the circumference subtending a $90^\circ$ arc is $45^\circ$.
Alternatively, the triangle formed by the diameter and two vertices is a right-angled isosceles triangle.
So the base angles are $45^\circ$.
Angle $p$ is one of these base angles?
Looking at the diagram, $p$ is the angle between the side and the diagonal.
Since the diagonals of a square bisect the $90^\circ$ corners, $p = 45^\circ$.
Also, the angle labeled $36^\circ$ is confusing. Wait, diagram 8 has a $36^\circ$ label.
If there is a $36^\circ$ label, it is NOT a square.
Let's re-examine Diagram 8.
Tick marks indicate equal length chords.
Top-Left to Top-Right has a tick.
Top-Left to Bottom-Left has a tick.
So Chord(TL-TR) = Chord(TL-BL).
This means the arcs are equal.
Consequently, the angles subtended by these arcs at the circumference are equal.
The angle $36^\circ$ is at the Top-Right vertex. It subtends the arc from Bottom-Right to Top-Left?
Let's trace the lines for the $36^\circ$ angle.
Vertex: Top-Right. Lines go to Bottom-Right and... Center? No, to Bottom-Left?
If the line goes to the center, it's a radius.
Let's assume the line with the dot is a diameter connecting Bottom-Left and Top-Right.
Then the angle at Top-Left is $90^\circ$ and angle at Bottom-Right is $90^\circ$.
The angle $36^\circ$ is at the Top-Right vertex.
It is formed by the diameter and the chord to the Bottom-Right?
If so, in the right-angled triangle (Top-Right, Bottom-Right, Bottom-Left), the angle at Bottom-Left would be $36^\circ$ (alternate segment? no).
Let's look at angle $p$.
Angle $p$ is at the Bottom-Left vertex.
It is formed by the diameter and the chord to the Top-Left.
We know Chord(TL-BL) = Chord(TL-TR).
Equal chords subtend equal angles at the circumference.
The angle subtended by Chord(TL-TR) at the circumference (e.g., at Bottom-Left) is angle $p$? No.
The angle subtended by Chord(TL-TR) at vertex Bottom-Left is $\angle(\text{BL-TL-TR})$? No.
Angle subtended by arc TR-TL at vertex BL is $\angle(\text{TL-BL-TR})$. This is angle $p$ IF the other line is the diameter.
The angle subtended by arc TL-BL at vertex TR is $\angle(\text{TL-TR-BL})$.
Since Chord(TL-TR) = Chord(TL-BL), Arc(TL-TR) = Arc(TL-BL).
Therefore, the angles subtended by these arcs at any point on the remaining circumference are equal.
Angle subtended by Arc(TL-BL) at vertex TR is the angle labeled $36^\circ$?
Let's check the lines for the $36^\circ$ angle.
Vertex TR. Lines to TL and BL?
If the lines are to TL and BL, then $36^\circ$ subtends Arc(TL-BL).
Angle $p$ is at vertex BL. Lines to TL and TR?
If the lines are to TL and TR, then $p$ subtends Arc(TL-TR).
Since Arc(TL-BL) = Arc(TL-TR), the angles are equal.
So $p = 36^\circ$.

9)
* Angle n: The reflex angle at the center is $240^\circ$.
The angle at the center inside the quadrilateral is $360^\circ - 240^\circ = 120^\circ$.
The triangle containing $n$ is formed by two radii and a chord. It is isosceles.
The vertex angle is $120^\circ$.
The base angles are $(180^\circ - 120^\circ) / 2 = 30^\circ$.
So $n = 30^\circ$.
*Correction*: Angle $n$ is marked as the whole angle at the bottom vertex of the cyclic quad?
No, $n$ is inside the triangle formed by the center and the bottom chord.
Wait, the lines for $n$ come from the bottom-left and bottom-right vertices meeting at the bottom vertex?
No, the vertex of angle $n$ is on the circle.
The angle at the center is $120^\circ$ (interior).
The angle at the circumference ($n$) subtending the same arc is half the center angle?
No, $n$ and the center angle are in the same segment?
Angle at center $= 120^\circ$.
Angle at circumference $n$ subtends the SAME arc?
If they subtend the same arc, $n = 120^\circ / 2 = 60^\circ$.
Let's verify the position.
Center angle faces the bottom chord.
Angle $n$ faces the bottom chord.
Yes, they are in the same segment.
So $n = 60^\circ$.
* Angle p: Angle $p$ is at the top right.
It subtends the arc on the left.
The central angle for that arc is...?
The total circle is $360^\circ$.
We have an isosceles triangle at the bottom with vertex angle $120^\circ$.
The tick marks indicate the bottom-left chord equals the bottom-right chord?
Ticks are on the chords from Bottom-Left to Center? No, ticks are on the chords of the quad?
Ticks are on the segments from the bottom vertex to the left/right vertices.
This implies the triangle with angle $n$ is isosceles?
If the chords are equal, the arcs are equal.
Let's assume symmetry.
If the setup is symmetric, the arc on the left equals the arc on the right.
Reflex angle $240^\circ$ corresponds to the major arc.
The minor arc (bottom) corresponds to $120^\circ$ at center.
The remaining $240^\circ$ is split between left and right arcs?
If the chords with ticks are equal, then Arc(Left) = Arc(Right).
So Arc(Left) $= 240^\circ / 2 = 120^\circ$.
Angle $p$ subtends Arc(Left)?
Angle $p$ is at the Top-Right vertex. It looks across to the Left arc?
The lines forming $p$ go to Top-Left and Bottom-Left.
So $p$ subtends Arc(Top-Left to Bottom-Left).
The central angle for this arc is $120^\circ$.
So angle at circumference $p = 120^\circ / 2 = 60^\circ$.
So $p = 60^\circ$.
* Angle q: Angle $q$ is opposite angle $p$? No.
Angle $q$ is at the Bottom-Right vertex.
It subtends the Top Arc?
The Top Arc is composed of Arc(Top-Left) + Arc(Top-Right)?
Actually, easier way: Cyclic Quadrilateral.
Opposite angles sum to $180^\circ$.
Which angle is opposite $q$? The Top-Left angle.
We don't have the Top-Left angle.
Let's find the angle opposite $p$. That is the Bottom-Left angle.
Bottom-Left angle $= n +$ part of the triangle?
Let's calculate the Top-Left angle.
It subtends the Right Arc ($120^\circ$). So Top-Left Angle $= 60^\circ$.
Opposite to Bottom-Right ($q$).
So $q + 60^\circ = 180^\circ \rightarrow q = 120^\circ$.
Let's double check.
Angle $q$ subtends the Major Arc (Left + Top + Right? No).
Angle $q$ (at Bottom-Right) subtends the arc from Top-Right to Top-Left to Bottom-Left.
Arc(Right) $= 120^\circ$. Arc(Top)?
Wait, the vertices are Top-Left, Top-Right, Bottom-Right, Bottom-Left.
Arc(Bottom) $= 120^\circ$.
Arc(Left) $= 120^\circ$.
Arc(Right) $= 120^\circ$.
Sum $= 360^\circ$. So Arc(Top) is effectively zero? No, there are 4 vertices.
My assumption that Arc(Left)=Arc(Right) was based on ticks.
Ticks are on Chord(BL-Center?) No, Chord(BL-TL) and Chord(BR-TR)?
Let's look at the ticks again.
Tick on Chord(Bottom-Left to Top-Left).
Tick on Chord(Bottom-Right to Top-Right).
This means Arc(Left) = Arc(Right).
We established Arc(Bottom) corresponds to central angle $120^\circ$.
So Arc(Left) + Arc(Right) $= 360^\circ - 120^\circ = 240^\circ$.
Since they are equal, Arc(Left) $= 120^\circ$, Arc(Right) $= 120^\circ$.
Angle $p$ (Top-Right) subtends Arc(Left).
Angle $= 120^\circ / 2 = 60^\circ$. Correct.
Angle $q$ (Bottom-Right) subtends Arc(Left) + Arc(Top)?
No, Angle $q$ is formed by chords to Top-Right and Bottom-Left.
It subtends Arc(Top-Right to Top-Left to Bottom-Left).
This arc is Arc(Right) + Arc(Top)? No.
The vertices are ordered TL, TR, BR, BL.
Angle at BR ($q$) subtends Arc(BL-TL-TR).
Arc(BL-TL) is Arc(Left) $= 120^\circ$.
Arc(TL-TR) is Arc(Top). What is Arc(Top)?
Wait, if Arc(Left)=120, Arc(Right)=120, Arc(Bottom)=120, then Arc(Top) must be 0?
This implies TL and TR are the same point? No.
Let's re-read the center angle.
Reflex angle $240^\circ$.
This is the angle covering Arc(Left) + Arc(Top) + Arc(Right).
So Arc(Left)+Arc(Top)+Arc(Right) $= 240^\circ$.
Arc(Bottom) $= 120^\circ$.
Ticks on Chord(TL-BL) and Chord(TR-BR).
So Arc(Left) = Arc(Right).
We don't know Arc(Top).
However, look at angle $n$.
$n$ is in triangle BL-Center-BR? No, $n$ is at the circumference.
Wait, earlier I said $n=60^\circ$.
Let's look at $p$ and $q$ again.
Angle $p$ (at TR) subtends Arc(BL-TL).
Angle $q$ (at BR) subtends Arc(TL-TR-BL)? No, Arc(TL-BL) + Arc(Top)?
Actually, $p$ and the angle at BL (let's call it $B_L$) subtend Arc(TR-BR)?
Let's use the property: Angle at center $= 2 \times$ Angle at circumference.
Angle $p$ subtends Arc(TL-BL).
Angle $q$ subtends Arc(TL-TR-BL)?
Let's find the angle at TL.
Angle at TL subtends Arc(TR-BR).
Arc(TR-BR) is Arc(Right).
We know Arc(Left) = Arc(Right).
Let Arc(Left) $= x$, Arc(Right) $= x$.
Arc(Bottom) $= 120^\circ$.
Arc(Top) $= y$.
$2x + y + 120 = 360 \rightarrow 2x + y = 240$.
We need more info.
Is the shape an isosceles trapezium?
If Arc(Left) = Arc(Right), then Chord(TL-BL) = Chord(TR-BR).
This makes it an isosceles trapezium if Top parallel to Bottom.
If so, Base angles are equal.
Angle(TL) = Angle(TR). Angle(BL) = Angle(BR).
Angle(BR) is $q$.
Angle(TR) is $p$? No, $p$ is part of the angle at TR?
Diagram 9: $p$ is the angle $\angle(\text{TL-TR-BL})$? No, $\angle(\text{TL-TR-BR})$?
The lines for $p$ are from TR to TL and TR to BL?
If so, $p$ subtends Arc(TL-BL) which is $x$.
So $p = x/2$.
The lines for $q$ are from BR to TR and BR to BL?
If so, $q$ subtends Arc(TR-TL-BL) which is $y + x$.
So $q = (x+y)/2$.
We know $2x + y = 240$.
We need $x$.
Is there a constraint I missed?
Maybe the triangle with $n$ is equilateral?
If Chord(BL-BR) corresponds to $120^\circ$, the triangle O-BL-BR is isosceles with $120^\circ$ vertex.
Ticks are on the OTHER chords.
Wait, look at angle $n$ again.
$n$ is the angle $\angle(\text{TL-BL-BR})$?
If $n$ subtends Arc(TL-TR-BR)?
Let's assume the standard case where the unmarked arc is determined by symmetry or specific values.
Often in these problems, if not specified, it might be a regular polygon? No.
Let's look at the solution for similar online problems.
Usually, $p$ and $q$ are related.
Let's guess $x=120$? If $x=120$, then $y=0$. Impossible.
Let's guess $y=120$? Then $2x=120, x=60$.
If $x=60$:
$p = 60/2 = 30^\circ$.
$q = (60+120)/2 = 90^\circ$.
Does this fit?
If $x=60$, Arc(Left)=60, Arc(Right)=60, Arc(Top)=120, Arc(Bottom)=120.
Check sum: $60+60+120+120=360$.
This creates an Isosceles Trapezium.
Is there evidence for $y=120$?
The angle $240^\circ$ is reflex.
Maybe the top triangle is equilateral?
Without explicit info, I will bet on the symmetry of the "top" part being similar to the "bottom" part or $p$ being simple.
However, looking at $p$, it looks like $30^\circ$ or $45^\circ$.
Looking at $q$, it looks like $90^\circ$ or $100^\circ$.
Let's try another path.
Angle $n$ is marked.
If $n$ subtends Arc(Top+Right)?
Let's assume the question implies $p=30^\circ, q=90^\circ$ based on $x=60$.

*Self-Correction*: Look at the ticks again.
Ticks on Chord(BL-TL) and Chord(BR-TR).
This confirms Arc(Left) = Arc(Right).
There is NO tick on Top or Bottom.
However, often "Calculate the missing angles" implies a unique solution.
Is it possible the center angle $240$ refers to the arc TL-TR-BR?
Yes, Reflex Angle.
So Arc(TL-TR-BR) $= 240^\circ$.
Arc(BL-TL) + Arc(TL-TR) + Arc(TR-BR) $= 240^\circ$.
We know Arc(BL-TL) = Arc(TR-BR) = $x$.
So $2x + \text{Arc(Top)} = 240$.
We still have 2 variables.
Is it possible $p$ and $q$ are not dependent on the split?
No.
Let's look at angle $n$ again.
$n$ is at BL. Lines to TL and BR.
Subtends Arc(TL-TR-BR).
Arc(TL-TR-BR) $= 240^\circ$.
So $n = 240 / 2 = 120^\circ$.
Wait, earlier I calculated $n=60$.
If $n=120$, then the triangle logic was wrong.
Let's re-evaluate $n$.
Angle $n$ is $\angle(\text{TL-BL-BR})$.
It subtends the Major Arc TL-TR-BR.
The measure of the inscribed angle is half the intercepted arc.
Intercepted Arc is $240^\circ$.
So $n = 120^\circ$.

Now, back to $p$ and $q$.
We still need the split of the $240^\circ$ arc.
Is there a visual clue?
The top triangle looks equilateral?
If Arc(Top) $= 120^\circ$, then $2x = 120 \rightarrow x=60$.
Then $p$ (subtends Arc(Left)=60) $= 30^\circ$.
$q$ (subtends Arc(Top)+Arc(Left) = $120+60=180$) $= 90^\circ$.
This seems the most likely intended "standard" geometry problem configuration.

So:
$n = 120^\circ$
$p = 30^\circ$
$q = 90^\circ$

10)
* Angle r: The triangle with the center has an angle of $16^\circ$ at the circumference? No, $16^\circ$ is at the bottom vertex.
The triangle is formed by Center, Bottom, Top-Right.
It is isosceles (radii).
So the angle at Top-Right (inside the triangle) is also $16^\circ$.
Angle $r$ is adjacent to this?
No, $r$ is the angle $\angle(\text{Top-Left-Top-Right-Bottom})$?
Let's trace $r$. Vertex Top-Right. Lines to Top-Left and Bottom.
Angle $r$ subtends Arc(Top-Left-Bottom).
We know Angle(Center-Bottom-Top-Right) $= 16^\circ$.
This is an isosceles triangle O-Bottom-TR.
Angle at Center O $= 180 - 16 - 16 = 148^\circ$.
This is the central angle for Arc(Bottom-TR).
We need Arc(TL-Bottom).
Look at the other triangle: O-Top-Left-Bottom?
We have angle $51^\circ$ at Top-Left.
Triangle O-TL-Bottom is isosceles.
Angle at TL is $51^\circ$? No, the $51^\circ$ is $\angle(\text{TR-TL-Bottom})$?
The lines for $51^\circ$ are from TL to TR and TL to Bottom.
So $51^\circ$ subtends Arc(TR-Bottom).
We just found Arc(TR-Bottom) corresponds to central angle $148^\circ$?
Inscribed angle $= 148 / 2 = 74^\circ$.
But the diagram says $51^\circ$. Contradiction.

Let's restart Diagram 10.
Angle $51^\circ$ is at Top-Left.
Angle $16^\circ$ is at Bottom.
Angle $s$ is at Top-Left (part of the angle?).
Angle $r$ is at Top-Right.

Let's assume the lines go to the center.
Triangle O-TL-Bottom: Isosceles.
Triangle O-Bottom-TR: Isosceles.
Angle $16^\circ$ is $\angle(\text{O-Bottom-TR})$?
If so, $\angle(\text{O-TR-Bottom}) = 16^\circ$.
Central Angle $\angle(\text{O-Bottom-TR}) = 180 - 32 = 148^\circ$.

Angle $51^\circ$ is $\angle(\text{TL-Bottom-O})$?
If the line goes to the center, then $\angle(\text{TL-Bottom-O}) = 51^\circ$.
Then $\angle(\text{O-TL-Bottom}) = 51^\circ$.
Central Angle $\angle(\text{O-TL-Bottom}) = 180 - 102 = 78^\circ$.

Now find $r$ and $s$.
$s$ is $\angle(\text{O-TL-TR})$?
$r$ is $\angle(\text{TL-TR-Bottom})$?

Let's look at $s$.
$s$ is part of the angle at TL.
The total angle at TL in the quad?
Actually, $s$ is labeled inside the triangle O-TL-TR?
If so, we need Central Angle $\angle(\text{O-TL-TR})$.
Total angle around center $= 360$.
We have $\angle(\text{O-TL-Bottom}) = 78^\circ$.
We have $\angle(\text{O-Bottom-TR}) = 148^\circ$.
Remaining angle $\angle(\text{O-TR-TL}) = 360 - 78 - 148 = 134^\circ$.
Triangle O-TR-TL is isosceles.
Base angles $= (180 - 134) / 2 = 23^\circ$.
So angle $\angle(\text{O-TL-TR}) = 23^\circ$.
Angle $\angle(\text{O-TR-TL}) = 23^\circ$.

Now identify $s$ and $r$.
$s$ is at TL. It looks like $\angle(\text{O-TL-Bottom}) + \angle(\text{O-TL-TR})$?
Or just $\angle(\text{O-TL-TR})$?
The arc for $s$ is small.
Let's assume $s = \angle(\text{O-TL-TR}) = 23^\circ$?
Or is $s$ the whole angle $\angle(\text{Bottom-TL-TR})$?
$\angle(\text{Bottom-TL-O}) = 51^\circ$.
$\angle(\text{O-TL-TR}) = 23^\circ$.
Total $s = 51 + 23 = 74^\circ$.

$r$ is at TR.
$\angle(\text{O-TR-Bottom}) = 16^\circ$.
$\angle(\text{O-TR-TL}) = 23^\circ$.
Total $r = 16 + 23 = 39^\circ$.

Let's check if $s$ and $r$ subtend arcs.
$s$ (Angle TL) subtends Arc(Bottom-TR).
Arc(Bottom-TR) central angle $= 148^\circ$.
Inscribed angle $= 148 / 2 = 74^\circ$. Matches.
$r$ (Angle TR) subtends Arc(TL-Bottom).
Arc(TL-Bottom) central angle $= 78^\circ$.
Inscribed angle $= 78 / 2 = 39^\circ$. Matches.

So:
$r = 39^\circ$
**$