In these problems the angles are given as expressions in $x$. You will need to use a circle theorem to form your own equation, solve it, and then find the angle asked for. There is a worked example to show you how to start; hints are at the foot of the sheet. No calculator.
Worked example: how to start
$A$, $B$ and $C$ are points on a circle. $AB$ is a diameter. Angle $CAB = 2x$ and angle $CBA = 3x$. Work out the value of $x$.
Because $AB$ is a diameter, angle $ACB = 90°$ (the angle in a semicircle).
The angles of triangle $ABC$ add to 180°: $2x+3x+90=180$.
$5x=90$, so $x=18$.
1$A$, $B$, $C$ and $D$ lie on a circle, forming a cyclic quadrilateral $ABCD$. Angle $A = 4x$ and the opposite angle $C = 5x$. Work out the size of angle $A$.[3 marks]
Answer:
2$A$, $B$ and $C$ are points on a circle, centre $O$. The angle $ABC$ at the circumference is $2x+5$. The angle $AOC$ at the centre, standing on the same arc $AC$, is $5x-20$. Work out the size of angle $ABC$.[4 marks]
Answer:
3$A$, $B$ and $C$ are points on a circle. The tangent to the circle at $A$ makes an angle of $3x-10$ with the chord $AB$. By the alternate segment theorem this equals the angle $ACB$ in the alternate segment, which is $2x+20$. Work out the size of angle $ACB$.[3 marks]
Answer:
4$A$ and $B$ are points on a circle, centre $O$. $OA$ and $OB$ are radii, and the angle $AOB$ at the centre is $2x$. The angle $OAB$ is $x+24$. Work out the value of $x$, and hence the size of angle $AOB$.[5 marks]
Answer:
5$ABCD$ is a cyclic quadrilateral. Angle $DAB = 2x+10$, angle $ABC = 3x$ and angle $BCD = x+50$. Work out the size of angle $ADC$.[5 marks]
Answer:
★Extension: $P$, $Q$ and $R$ are points on a circle, centre $O$. Angle $PQR = y$. Explain why the reflex angle $POR$ at the centre is $2y$.
Stuck? Hints (don't peek unless you need to)1. Opposite angles of a cyclic quadrilateral add to 180°.2. The angle at the centre is twice the angle at the circumference when both stand on the same arc.3. The two expressions describe the SAME angle (alternate segment theorem), so set them equal.4. Triangle $OAB$ is isosceles because $OA$ and $OB$ are both radii, so its two base angles are equal. Use the angle sum of the triangle.5. Opposite angles sum to 180°. Use the pair $A$ and $C$ to find $x$ first, then use the pair $B$ and $D$.
Solutions & mark scheme · Circle Theorems
Total: 20 marks
Award the marks shown for each correct step; many of these have more than one valid route, so give method marks for any correct working.
1$A$, $B$, $C$ and $D$ lie on a circle, forming a cyclic quadrilateral $ABCD$. Angle $A = 4x$ and the opposite angle $C = 5x$. Work out the size of angle $A$.[3]
Model solution
Opposite angles add to 180°: $4x+5x=180$.
$9x=180$, so $x=20$.
Angle $A=4x=4\times20=80°$.
Answer: 80°
Marks
✔1$4x+5x=180$
✔1$x=20$
✔1$80°$
2$A$, $B$ and $C$ are points on a circle, centre $O$. The angle $ABC$ at the circumference is $2x+5$. The angle $AOC$ at the centre, standing on the same arc $AC$, is $5x-20$. Work out the size of angle $ABC$.[4]
Model solution
Angle at centre $=2\times$ angle at circumference: $5x-20=2(2x+5)$.
$5x-20=4x+10$, so $x=30$.
Angle $ABC=2x+5=2\times30+5=65°$.
Answer: 65°
Marks
✔1$5x-20=2(2x+5)$
✔1$5x-20=4x+10$
✔1$x=30$
✔1$65°$
3$A$, $B$ and $C$ are points on a circle. The tangent to the circle at $A$ makes an angle of $3x-10$ with the chord $AB$. By the alternate segment theorem this equals the angle $ACB$ in the alternate segment, which is $2x+20$. Work out the size of angle $ACB$.[3]
Model solution
Alternate segment theorem: $3x-10=2x+20$.
So $x=30$.
Angle $ACB=2x+20=2\times30+20=80°$.
Answer: 80°
Marks
✔1$3x-10=2x+20$
✔1$x=30$
✔1$80°$
4$A$ and $B$ are points on a circle, centre $O$. $OA$ and $OB$ are radii, and the angle $AOB$ at the centre is $2x$. The angle $OAB$ is $x+24$. Work out the value of $x$, and hence the size of angle $AOB$.[5]
Model solution
$OA=OB$ (radii), so triangle $OAB$ is isosceles and the base angles are equal: angle $OBA=$ angle $OAB=x+24$.
Angle sum of the triangle: $2x+(x+24)+(x+24)=180$.
$4x+48=180$, so $4x=132$ and $x=33$.
Angle $AOB=2x=2\times33=66°$.
Answer: x = 33, angle AOB = 66°
Marks
✔1$OA=OB$, so base angles equal
✔1Both base angles $=x+24$
✔1$2x+2(x+24)=180$
✔1$x=33$
✔1angle $AOB=66°$
5$ABCD$ is a cyclic quadrilateral. Angle $DAB = 2x+10$, angle $ABC = 3x$ and angle $BCD = x+50$. Work out the size of angle $ADC$.[5]
Model solution
Opposite angles $DAB$ and $BCD$ add to 180°: $(2x+10)+(x+50)=180$.
$3x+60=180$, so $x=40$.
Angle $ABC=3x=120°$.
Angles $ABC$ and $ADC$ are opposite, so add to 180°: angle $ADC=180-120=60°$.
Answer: 60°
Marks
✔1$DAB+BCD=180$
✔1$3x+60=180$
✔1$x=40$
✔1angle $ABC=120°$
✔1angle $ADC=60°$
★Extension
The angle at the centre is twice the angle at the circumference when both stand on the same arc.
Angle $PQR$ stands on the arc $PR$ that does NOT contain $Q$.
The angle at the centre standing on that same arc is the REFLEX angle $POR$ (it opens towards $Q$).
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