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 $D$ are four points on a circle, in that order, forming a cyclic quadrilateral $ABCD$. Angle $DAB = 4x+5$ and the opposite angle $BCD = x+65$. A student argues that opposite angles of a cyclic quadrilateral are always equal to each other, so writes $4x+5=x+65$ and gets $x=20$. State what is wrong with the student's method, then use the correct theorem to work out the true size of angle $DAB$.[3 marks]
Answer:
2$P$ is a point outside a circle, centre $O$. $PA$ and $PB$ are tangents to the circle, touching it at $A$ and $B$. Because $PA=PB$, triangle $PAB$ is isosceles, so angle $PAB=$ angle $PBA$. Angle $APB = x^2+x+50$ and angle $PAB = 3x+5$. Work out the size of angle $APB$, rejecting any value of $x$ that does not give a valid triangle.[4 marks]
Answer:
3$A$, $B$ and $C$ are points on a circle, centre $O$. The angle $ABC$ at the circumference is $25$ degrees. The angle $AOC$ at the centre, standing on the same arc $AC$, is $5x$, where $x$ is a positive integer. Work out the value of $x$.[3 marks]
Answer:
4$A$, $B$, $C$ and $D$ are four points on a circle, in that order around the circle, and $AB$ is a diameter. $ABCD$ is a cyclic quadrilateral. Angle $CAB = x$ and angle $ADC = 4x-30$. Work out the size of angle $ABC$.[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 sum to 180 degrees; they are not equal to each other.2. Form the equation from the triangle's angle sum, solve the resulting quadratic, then check which root gives sensible angles.3. Use the centre-circumference theorem first to find the size of angle AOC itself, then use that to find x.4. Find angle ABC one way using the semicircle theorem and the triangle angle sum, and another way using the cyclic quadrilateral, then combine the two.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 $D$ are four points on a circle, in that order, forming a cyclic quadrilateral $ABCD$. Angle $DAB = 4x+5$ and the opposite angle $BCD = x+65$. A student argues that opposite angles of a cyclic quadrilateral are always equal to each other, so writes $4x+5=x+65$ and gets $x=20$. State what is wrong with the student's method, then use the correct theorem to work out the true size of angle $DAB$.[3]
Model solution
The student is wrong: opposite angles of a cyclic quadrilateral sum to 180 degrees, they are not equal.
Correct equation: $4x+5+x+65=180$, so $5x+70=180$.
$5x=110$, so $x=22$.
Angle $DAB=4\times22+5=93$ degrees.
Answer: x = 22, angle DAB = 93 degrees
Marks
✔1Identifies the correct rule (opposite angles sum to 180, not equal) and sets up $4x+5+x+65=180$
✔1$x=22$
✔1Angle $DAB=93$ degrees
2$P$ is a point outside a circle, centre $O$. $PA$ and $PB$ are tangents to the circle, touching it at $A$ and $B$. Because $PA=PB$, triangle $PAB$ is isosceles, so angle $PAB=$ angle $PBA$. Angle $APB = x^2+x+50$ and angle $PAB = 3x+5$. Work out the size of angle $APB$, rejecting any value of $x$ that does not give a valid triangle.[4]
Model solution
Angle sum of triangle $PAB$: $(x^2+x+50)+2(3x+5)=180$.
This simplifies to $x^2+7x-120=0$, which factorises as $(x-8)(x+15)=0$, so $x=8$ or $x=-15$.
Reject $x=-15$: it gives angle $PAB=3(-15)+5=-40$ degrees, which is impossible for a triangle. Keep $x=8$.
Angle $APB=8^2+8+50=122$ degrees.
Answer: x = 8, angle APB = 122 degrees (x = -15 is rejected)
Marks
✔1Correct equation from the angle sum, $(x^2+x+50)+2(3x+5)=180$, simplified to $x^2+7x-120=0$
✔1Solves to $x=8$ or $x=-15$
✔1Rejects $x=-15$ with a reason (negative or impossible angle), keeps $x=8$
✔1Angle $APB=122$ degrees
3$A$, $B$ and $C$ are points on a circle, centre $O$. The angle $ABC$ at the circumference is $25$ degrees. The angle $AOC$ at the centre, standing on the same arc $AC$, is $5x$, where $x$ is a positive integer. Work out the value of $x$.[3]
Model solution
Angle at the centre is twice angle at the circumference: angle $AOC=2\times25=50$ degrees.
So $5x=50$.
$x=10$.
Answer: x = 10
Marks
✔1Angle $AOC=2\times25=50$ degrees
✔1Equation $5x=50$
✔1$x=10$
4$A$, $B$, $C$ and $D$ are four points on a circle, in that order around the circle, and $AB$ is a diameter. $ABCD$ is a cyclic quadrilateral. Angle $CAB = x$ and angle $ADC = 4x-30$. Work out the size of angle $ABC$.[5]
Model solution
$AB$ is a diameter, so angle $ACB=90$ degrees (angle in a semicircle).
Angle sum of triangle $ABC$: angle $ABC=180-90-x=90-x$.
$ABCD$ is a cyclic quadrilateral, so angle $ABC$ and angle $ADC$ are opposite and sum to 180 degrees: angle $ABC=180-(4x-30)=210-4x$.
Setting the two routes equal: $90-x=210-4x$, so $3x=120$ and $x=40$.
Angle $ABC=90-40=50$ degrees.
Answer: angle ABC = 50 degrees
Marks
✔1Angle $ACB=90$ degrees (angle in a semicircle)
✔1Angle $ABC=90-x$ from the triangle angle sum
✔1Angle $ABC=210-4x$ from the cyclic quadrilateral (opposite angles sum to 180)
✔1$90-x=210-4x$ solved to $x=40$
✔1Angle $ABC=50$ degrees
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 $Q$).
Circle Theorems