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

Here are the step-by-step solutions for each problem on the worksheet.

1)
* a) Angles in the same segment are equal. The angle opposite $93^\circ$ is also $93^\circ$.
* b) Opposite angles in a cyclic quadrilateral add up to $180^\circ$. So, $b + 93^\circ = 180^\circ$. Therefore, $b = 180 - 93 =$ $87^\circ$.

2)
* c) Angles in the same segment are equal. The angle opposite $69^\circ$ is also $69^\circ$.
* d) Angles on a straight line add up to $180^\circ$. The interior angle next to $102^\circ$ is $180 - 102 = 78^\circ$. In a cyclic quadrilateral, opposite angles sum to $180^\circ$. So, $d + 78^\circ = 180^\circ$. Therefore, $d = 180 - 78 =$ $102^\circ$. (Alternatively, the exterior angle of a cyclic quad equals the interior opposite angle).

3)
* e) The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. So, $e$ is equal to the angle at the top left ($88^\circ$). Answer: $88^\circ$.
* f) First, find the interior angle at the top right. It forms a linear pair with $91^\circ$, so it is $180 - 91 = 89^\circ$. Opposite angles sum to $180^\circ$, so $f + 89^\circ = 180^\circ$. Therefore, $f = 180 - 89 =$ $91^\circ$.

4)
* g) The triangle contains the center of the circle, making the hypotenuse a diameter. The angle subtended by a diameter at the circumference is always $90^\circ$. So, $g =$ $90^\circ$.
* h) The angles in a triangle add up to $180^\circ$. We have $90^\circ$ and $37^\circ$. So, $h = 180 - 90 - 37 =$ $53^\circ$.

5)
* i) The angle at the center is double the angle at the circumference subtended by the same arc. The angle at the circumference is $70^\circ$. So, $j = 2 \times 70 = 140^\circ$. Angle $i$ and angle $j$ are on a straight line (the diameter). So, $i = 180 - 140 =$ $40^\circ$.
* *Alternative method:* The triangle containing $i$ is isosceles (two radii). The base angles are equal. The central angle for that specific triangle sector is $180 - 140 = 40$? No, simpler way: The angle at the center subtending the arc for the $70^\circ$ angle is $140^\circ$. The remaining angle on the straight line diameter is $180-140=40^\circ$. This is the central angle for the triangle containing $i$. Since the triangle is isosceles (radii), the base angles are $(180-40)/2 = 70^\circ$. Wait, looking at diagram 5 again.
* Let's re-evaluate 5 carefully.
* Angle at circumference = $70^\circ$.
* Angle at center ($j$) = $2 \times 70^\circ = 140^\circ$.
* Angle $i$ is part of an isosceles triangle formed by two radii and a chord. The central angle of this triangle is adjacent to $j$ on the straight diameter line. So the central angle is $180^\circ - 140^\circ = 40^\circ$.
* The triangle with angle $i$ has two equal sides (radii). So the base angles are equal. $i = (180^\circ - 40^\circ) / 2 = 140 / 2 =$ $70^\circ$.
* Actually, there is a theorem: Angles in the same segment are equal. If we drew a chord from the end of the diameter to the vertex of the $70^\circ$ angle, we'd see relationships. But simpler: The triangle containing $i$ and the triangle containing the $70^\circ$ angle share the same circumcircle properties.
* Let's stick to the calculation: Central angle reflex $j$? No, $j$ is obtuse. $j = 140^\circ$. The angle adjacent to $j$ on the diameter is $40^\circ$. The triangle containing $i$ is isosceles with vertex angle $40^\circ$. Base angles $i = (180-40)/2 = 70^\circ$.
* Correction: Looking closely at diagram 5, $i$ and the $70^\circ$ angle are in the *same segment* if we consider the chord connecting their endpoints? No.
* Let's look at the triangle with angle $i$. It is a right-angled triangle because it sits on the diameter? No, the vertex is not on the diameter.
* Let's use the property: Angle at center = $2 \times$ Angle at circumference.
* Angle $j = 2 \times 70^\circ = 140^\circ$.
* Angles on a straight line: The angle next to $j$ is $180^\circ - 140^\circ = 40^\circ$.
* The triangle containing angle $i$ has a central angle of $40^\circ$ and two sides that are radii (equal length). Therefore, it is an isosceles triangle.
* The base angles are equal: $i = (180^\circ - 40^\circ) \div 2 = 70^\circ$.
* So, $i = 70^\circ$.
* And $j = 140^\circ$.

6)
* k) The angle subtended by a diameter at the circumference is $90^\circ$. So, $k = 90^\circ$.
* l) Angles in the same segment are equal. The angle opposite $51^\circ$ is also $51^\circ$.

7)
* m) The tick marks indicate the two sides are equal lengths (chords). Equal chords subtend equal angles at the circumference. However, looking at the triangle containing $m$, it's not immediately obvious. Let's look at the quadrilateral.
* Actually, simpler approach: The triangle with the $67^\circ$ and $76^\circ$ angles... wait, those are angles of the quadrilateral vertices.
* Let's look at the triangle formed by the diagonal.
* Angle $n$ and the angle labeled $67^\circ$ are in the same segment? No.
* Let's use the property of parallel lines or isosceles triangles? The tick marks are on the top side and the right side. This means Chord Top = Chord Right.
* Therefore, the arcs they cut are equal.
* Therefore, the angles subtended by these arcs at the circumference are equal.
* The angle subtended by the Top Chord is $n$ (if we draw a diagonal from bottom-left to top-right). No, $n$ is subtended by the Right Chord? No, $n$ is an angle inside the triangle.
* Let's look at the angles subtended by the equal chords at the center? No center shown.
* Let's look at the angles at the circumference. The angle subtended by the "Top" chord at the bottom-left vertex is part of the $76^\circ$ angle? No.
* Let's assume the tick marks mean the triangle containing $m$ and $n$ is isosceles? No, the ticks are on the quadrilateral sides.
* Okay, Chord(Top) = Chord(Right).
* Angle subtended by Chord(Top) at the bottom-left vertex = Angle subtended by Chord(Right) at the bottom-left vertex? No.
* Angle subtended by Chord(Top) at the bottom-right vertex ($m$) = Angle subtended by Chord(Right) at the top-left vertex ($67^\circ$?? No, $67$ is the whole corner).
* Let's try this: Draw a diagonal from top-left to bottom-right.
* Angle subtended by Chord(Top) at bottom-right is $m$.
* Angle subtended by Chord(Right) at top-left is... let's call it $y$.
* Since Chords are equal, Arcs are equal. So Angle(Top-Left, subtending Right Chord) = Angle(Bottom-Right, subtending Top Chord).
* So, the angle $\angle$(Top-Left diagonal to Right Side) = $m$.
* We know the whole angle at Top-Left is $67^\circ$.
* We need more info.
* Let's look at the other diagonal.
* Angle subtended by Chord(Top) at Bottom-Left = Angle subtended by Chord(Right) at Top-Right.
* Let's look at the triangle with $76^\circ$.
* Actually, usually in these problems, if two chords are equal, the trapezoid is isosceles or similar.
* Let's look at angle $n$. $n$ is in the same segment as the angle at the top left? No.
* Let's calculate the third angle in the triangle on the left? We don't have enough angles.
* Let's re-read the diagram. Tick marks on Top Side and Right Side.
* This implies Arc(Top) = Arc(Right).
* Therefore, the inscribed angles subtending these arcs are equal.
* Angle subtending Arc(Top) is $\angle$(Bottom-Left to Top-Right diagonal)? No, it's $\angle$(Bottom-Right vertex looking at Top side). That is angle $m$.
* Angle subtending Arc(Right) is $\angle$(Top-Left vertex looking at Right side). Let's call the diagonal TL-BR. The angle between TL-diagonal and TL-TopSide? No. The angle is $\angle$(Top-Left vertex, specifically the part subtending the Right side).
* Let's denote vertices TL, TR, BR, BL.
* Chord TL-TR = Chord TR-BR.
* Angle $\angle$TBL (subtending TL-TR) = Angle $\angle$LTR (subtending TR-BR)? No.
* Angle $\angle$TBR (subtending TL-TR) ... wait.
* Angle at circumference subtending Arc(TL-TR) is $\angle$TBL? No, $\angle$TBR? No. It is $\angle$T(BL)R? No.
* The angle subtended by chord TL-TR at the circumference is $\angle$T(BL)R? No, it's $\angle$T(BR)L? No.
* It is $\angle$T(BL)R? No. The vertices are T, R, B, L.
* Chord TR. Angle at L is $\angle$TLR. Angle at B is $\angle$TBR.
* Chord RB. Angle at T is $\angle$RTB? No, $\angle$R(T)L? No. Angle at L is $\angle$RLB? No.
* Angle subtended by Chord RB at T is $\angle$RTB? No, $\angle$R(TopLeft)B? Yes, $\angle$LTB? No. $\angle$ATB where A is top left. Let's call TopLeft A, TopRight B, BotRight C, BotLeft D.
* Chord AB = Chord BC.
* Angle subtended by AB at D is $\angle$ADB.
* Angle subtended by BC at A is $\angle$BAC.
* So $\angle$ADB = $\angle$BAC.
* Also Angle subtended by AB at C is $\angle$ACB.
* Angle subtended by BC at D is $\angle$BDC.
* So $\angle$ACB = $\angle$BDC.
* We are given $\angle$DAB = $67^\circ$ and $\angle$ADC = $76^\circ$.
* We need $m$ ($\angle$ACB? No, $m$ is $\angle$BCA? Or $\angle$ACD? Diagram shows $m$ is $\angle$ACB).
* And $n$ is $\angle$CAD? Or $\angle$ADB? Diagram shows $n$ is $\angle$ADB.
* From above: $\angle$ADB ($n$) = $\angle$BAC.
* And $\angle$ACB ($m$) = $\angle$BDC.
* Also, since AB=BC, Triangle ABC is isosceles? No, we don't know AC. But Arc AB = Arc BC.
* Therefore, Chord AC bisects nothing necessarily.
* However, $\angle$BAC = $\angle$BCA? No. $\angle$BAC subtends Arc BC. $\angle$BCA subtends Arc AB. Since Arc AB = Arc BC, then $\angle$BAC = $\angle$BCA.
* So in Triangle ABC, $\angle$BAC = $\angle$BCA = $m$.
* Also $\angle$ADB ($n$) subtends Arc AB. So $n = m$.
* So $n = m$.
* Now look at Triangle ADC.
* Sum of angles = $180^\circ$.
* $\angle$DAC + $\angle$ADC + $\angle$ACD = $180^\circ$.
* We know $\angle$ADC = $76^\circ$.
* $\angle$DAB = $67^\circ$. $\angle$DAB = $\angle$DAC + $\angle$CAB = $\angle$DAC + $m$.
* So $\angle$DAC = $67 - m$.
* $\angle$BCD = $\angle$BCA + $\angle$ACD = $m + \angle$ACD.
* Opposite angles in cyclic quad sum to $180^\circ$.
* $\angle$DAB + $\angle$BCD = $180^\circ \Rightarrow 67 + \angle$BCD = $180 \Rightarrow \angle$BCD = $113^\circ$.
* So $m + \angle$ACD = $113 \Rightarrow \angle$ACD = $113 - m$.
* Now back to Triangle ADC sum:
* $(67 - m) + 76 + (113 - m) = 180$.
* $256 - 2m = 180$.
* $2m = 76$.
* $m = 38^\circ$.
* Since $n = m$, $n = 38^\circ$.
* So, $m = 38^\circ$, $n = 38^\circ$.

8)
* p) The quadrilateral has tick marks on all four sides, meaning it is a rhombus (or square). A rhombus inscribed in a circle must be a square.
* Therefore, all corner angles are $90^\circ$.
* The diagonal bisects the corner angle.
* So, $p = 90 / 2 =$ $45^\circ$.
* Alternatively: The triangle containing $p$ and $36^\circ$? No, $36^\circ$ is elsewhere.
* Wait, let's look at diagram 8 again.
* Tick marks on Left, Top, Right, Bottom? No.
* Tick marks on Left and Top. And a dot in the center.
* Wait, the tick marks are on the Left side and the Top side.
* And there is a diagonal drawn.
* Angle given is $36^\circ$. This angle is $\angle$(Top-Right vertex, between diagonal and Right side).
* Let vertices be TL, TR, BR, BL. Center O.
* Given: Chord TL-TR = Chord TL-BL? No, ticks are on TL-TR and BL-TL?
* Let's assume standard notation: Ticks on adjacent sides TL-TR and TL-BL? Or TL-TR and TR-BR?
* Looking at the image: Ticks are on the Top-Left side and the Bottom-Left side? No.
* Ticks are on the side connecting ~10 o'clock and ~2 o'clock (Top) and ~10 o'clock and ~8 o'clock (Left).
* So Chord Top = Chord Left.
* This means Arc Top = Arc Left.
* Angle $p$ is at the Bottom-Left vertex? No, $p$ is at the Bottom-Left vertex, inside the triangle formed by the diagonal from Top-Right to Bottom-Left?
* No, the diagonal is from Top-Left to Bottom-Right?
* Let's trace the lines. There is a diagonal from Top-Left to Bottom-Right.
* Angle $36^\circ$ is at the Top-Right vertex, between the diagonal and the Right side.
* Angle $p$ is at the Bottom-Left vertex, between the diagonal and the Left side.
* These two angles ($36^\circ$ and $p$) subtend the same arc?
* Angle $36^\circ$ ($\angle$TR-Diag-BR) subtends Arc BR-DiagEnd? No.
* Let Diagonal be DB (Top-Left to Bottom-Right).
* Angle $36^\circ$ is $\angle$TR-DB-BR? No, it's $\angle$TR-DB-RightSide?
* Vertex is Top-Right. Sides are Diagonal and Right-Chord.
* So it subtends the Arc from Bottom-Right to Top-Left? No.
* Angle $\angle$(TopRight) subtends Arc(BottomRight to TopLeft).
* Angle $p$ is at Bottom-Left. Sides are Diagonal and Left-Chord.
* So it subtends Arc(TopLeft to TopRight)? No.
* Angle $\angle$(BottomLeft) subtends Arc(TopRight to TopLeft).
* So Angle $36^\circ$ and Angle $p$ subtend the SAME ARC (The Top Arc).
* Therefore, $p = 36^\circ$.
* (The tick marks might be extra info or implying symmetry, but "Angles in the same segment" solves it directly).

9)
* n) The angle at the center is $240^\circ$ (reflex). The angle at the center inside the quadrilateral is $360 - 240 = 120^\circ$.
* The triangle containing $n$ is formed by two radii. It is isosceles.
* Vertex angle is $120^\circ$.
* Base angles $n = (180 - 120) / 2 = 30^\circ$.
* So, $n = 30^\circ$.
* p) Angle at circumference is half angle at center.
* The angle at the center subtending the same arc as $p$ is the reflex angle $240^\circ$? No.
* $p$ subtends the minor arc. The central angle for the minor arc is $120^\circ$.
* So $p = 120 / 2 = 60^\circ$.
* So, $p = 60^\circ$.
* q) Opposite angles in a cyclic quad sum to $180^\circ$.
* The quad is formed by the two radii and the two chords? No, the vertices are on the circle.
* Vertices: Top, Right, Bottom, Left.
* Angle at Left is $p = 60^\circ$.
* Angle at Right is $q$.
* Wait, is it a cyclic quad? Yes, all vertices on circle.
* So $p + q = 180^\circ$.
* $60 + q = 180 \Rightarrow q = 120^\circ$.
* So, $q = 120^\circ$.

10)
* r) Angle at center is double angle at circumference.
* Angle at circumference is $16^\circ$.
* So the central angle adjacent to the one shown is $2 \times 16 = 32^\circ$.
* Wait, $r$ is an angle at the circumference.
* $r$ and $51^\circ$ are in the same segment? No.
* Let's find the central angles.
* Angle subtended by the bottom chord at circumference is $16^\circ$. So central angle is $32^\circ$.
* Angle subtended by the left chord at circumference is $51^\circ$. So central angle is $102^\circ$.
* The remaining central angle (for the top chord) is $360 - 102 - 32 = 226^\circ$? No, that's the reflex.
* Let's check the position.
* $51^\circ$ is $\angle$(Top-Left, Bottom-Left, Top-Right)? No.
* Let vertices be TL, TR, BR, BL. Center O.
* $51^\circ$ is $\angle$TL-BL-TR? No, looks like $\angle$O-BL-TL? No.
* It is $\angle$TL-BL-TR? No, the line goes to the center.
* Ah, the lines go to the CENTER.
* So $51^\circ$ is NOT an inscribed angle. It is $\angle$O-BL-TL?
* Triangle O-BL-TL is isosceles (radii).
* If $51^\circ$ is the base angle $\angle$O-BL-TL, then $\angle$O-TL-BL is also $51^\circ$.
* Then central angle $\angle$TOL = $180 - 51 - 51 = 78^\circ$.
* Similarly, $16^\circ$ is $\angle$O-BL-BR?
* Triangle O-BL-BR is isosceles. Base angles $16^\circ$.
* Central angle $\angle$BOR = $180 - 16 - 16 = 148^\circ$.
* Then central angle $\angle$TOR (top) = $360 - 78 - 148 = 134^\circ$.
* Triangle O-TR-TL is isosceles.
* Angle $r$ is $\angle$O-TR-TL? Or $\angle$O-TR-BR?
* Diagram shows $r$ is $\angle$O-TR-TL? No, $r$ is the angle between the radius and the chord TR-TL?
* Actually, $r$ is marked as the angle $\angle$O-TR-TL.
* In Triangle O-TR-TL, vertex angle is $134^\circ$.
* Base angles $r = (180 - 134) / 2 = 23^\circ$.
* So, $r = 23^\circ$.
* s) $s$ is the whole angle at TL? No, $s$ is $\angle$O-TL-BL + $\angle$O-TL-TR?
* $\angle$O-TL-BL = $51^\circ$ (base angle of first triangle).
* $\angle$O-TL-TR = $23^\circ$ (base angle of top triangle).
* $s = 51 + 23 = 74^\circ$.
* So, $s = 74^\circ$.

11)
* t) The triangle has two sides as radii (tick marks). It is isosceles.
* The base angles are equal. One is $47^\circ$. So the other base angle is $47^\circ$.
* The vertex angle (at center) is $180 - 47 - 47 = 86^\circ$.
* $t$ is the angle at the circumference subtending the same arc?
* No, $t$ is the other base angle of the OTHER isosceles triangle?
* Let's look at the shape. It's a kite or two triangles sharing a radius?
* No, it's a quadrilateral with center O.
* Left Triangle: Isosceles. Base angles $47^\circ$. Central Angle = $86^\circ$.
* Right Triangle: Has tick marks on the two radii? Yes. And tick mark on the chord?
* The tick marks are on the Left Radius, Right Radius, and Right Chord?
* No, ticks are on Left Chord and Right Chord?
* Let's assume the ticks mean Chord Left = Chord Right.
* If Chords are equal, Central Angles are equal.
* So Right Central Angle = $86^\circ$.
* Right Triangle is isosceles with vertex $86^\circ$.
* Base angles $u = (180 - 86) / 2 = 47^\circ$.
* So $u = 47^\circ$.
* What is $t$? $t$ is the angle at the top vertex of the Right Triangle?
* Yes, $t$ corresponds to $u$'s partner?
* In the right triangle, the angles are $u$, $u$, and central.
* $t$ is marked as the top angle. So $t = u = 47^\circ$.
* So, $t = 47^\circ$, $u = 47^\circ$.

12)
* v) Angles in the same segment are equal.
* Angle $v$ and angle $29^\circ$ both subtend the same small arc at the top.
* So, $v = 29^\circ$.
* w) Angles in the same segment are equal.
* Angle $w$ and angle $41^\circ$? No.
* Angle $w$ and angle $x$?
* Let's find $x$ first.
* Angle $87^\circ$ and angle $w+41^\circ$?
* Look at the triangle with $87^\circ$ and $29^\circ$.
* Third angle is $180 - 87 - 29 = 64^\circ$.
* This $64^\circ$ angle is vertically opposite to the angle in the bottom triangle?
* Or, angle $87^\circ$ subtends the bottom-right arc.
* Angle $w+41^\circ$ subtends the top-left arc?
* Let's use "Angles in same segment".
* Angle $87^\circ$ is $\angle$(Left, Top, Right)? No.
* Let vertices be L, T, R, B.
* $\angle$LTR = ?
* $\angle$TLR = $87^\circ$? No, $87$ is at L. $\angle$TLR? No, $\angle$BLR?
* Diagram: Angle at Left vertex is split into $87^\circ$ (top part) and... wait.
* The angle $87^\circ$ is $\angle$T-L-R? No, the line goes to R.
* So $\angle$T-L-R = $87^\circ$.
* Angle at Top vertex is split into $29^\circ$ (right part) and...
* So $\angle$T-R-L?
* Let's identify segments.
* Chord TR. Angle at L is $\angle$TLR. Angle at B is $\angle$TBR.
* We don't have $\angle$TLR.
* We have $\angle$L-T-R? No.
* Let's look at the intersection X.
* Triangle Top-Left-X: Angles are $87^\circ$ (at L), $29^\circ$ (at T).
* So Angle at X = $180 - 87 - 29 = 64^\circ$.
* Vertically opposite angle at X (in Bottom-Right triangle) is $64^\circ$.
* Triangle Bottom-Right-X: Angles are $64^\circ$ (at X), $41^\circ$ (at B? No, at R? No).
* Angle $41^\circ$ is at the Bottom vertex? No, it's at the Right vertex?
* Label $41^\circ$ is near the bottom-right vertex. It is $\angle$X-R-B? Or $\angle$X-B-R?
* It looks like $\angle$X-B-R (Angle at Bottom).
* And $w$ is $\angle$X-R-B (Angle at Right).
* So in Triangle X-B-R:
* Angle X = $64^\circ$.
* Angle B = $41^\circ$.
* Angle R ($w$) = $180 - 64 - 41 = 75^\circ$.
* So, $w = 75^\circ$.
* x) $x$ is the angle at the intersection? No, $x$ is $\angle$L-X-R?
* $x$ and the $64^\circ$ angle are on a straight line?
* $x$ is the angle $\angle$L-X-R.
* Angles on straight line TR? No.
* $x$ is vertically opposite to the angle in the Left-Bottom triangle?
* Actually, $x$ is the exterior angle to the Top-Left triangle?
* $x = \angle$L-X-R.
* $\angle$L-X-T = $64^\circ$.
* $\angle$L-X-R and $\angle$L-X-T are supplementary? Yes, on straight line TR.
* So $x = 180 - 64 = 116^\circ$.
* So, $x = 116^\circ$.

Final Answer:
1) a) $93^\circ$, b) $87^\circ$
2) c) $69^\circ$, d) $102^\circ$
3) e) $88^\circ$, f) $91^\circ$
4) g) $90^\circ$, h) $53^\circ$
5) i) $70^\circ$, j) $140^\circ$
6) k) $90^\circ$, l) $51^\circ$
7) m) $38^\circ$, n) $38^\circ$
8) p) $36^\circ$
9) n) $30^\circ$, p) $60^\circ$, q) $120^\circ$
10) r) $23^\circ$, s) $74^\circ$
11) t) $47^\circ$, u) $47^\circ$
12) v) $29^\circ$, w) $75^\circ$, x) $116^\circ$