Faded examples · Finding angles with circle theorems
Non-calculator
Each worked example uses a different circle theorem, and shows a little less working than the one before – you complete the faded (blank) steps yourself, following the same four steps every time. In each example also answer the check question – it tests why the step works. No calculator, and always state your reason.
①Example 1fully worked: read it through
AB is a diameter. Angle CAB $=35°$. Find angle $x=$ angle CBA.
1Name the circle theorem
AB is a diameter, so the angle in a semicircle is a right angle
2Write the fact it gives you
angle ACB $=90°$ – mark it at C
3Set up the calculation
triangle ACB: $35°+90°+x=180°$
4Work out $x$
$x=180°-125°=55°$
Check · Why is angle ACB exactly $90°$?
A because AB is a diameter, and the angle in a semicircle is a right angleB because triangle ACB is equilateralC because $35°$ doubled is $70°$D because every angle in a circle is $90°$
②Example 2you finish the last 1 step
$A$, $B$ and $C$ are points on a circle, centre $O$. Angle $AOB=96°$. Find angle $x=$ angle $ACB$.
1Name the circle theorem
The angle at the centre is twice the angle at the circumference
2Write the fact it gives you
angle $AOB$ (at the centre) and angle $x$ (at $C$) stand on the same arc $AB$
3Set up the calculation
$x = 96° \div 2$
4Work out $x$
$x = 48°$
Check · Angle $AOB$ is at the centre. How is it related to angle $x$ at the circumference?
A they are equalB they add up to $180°$C the centre angle is twice the circumference angleD the centre angle is half the circumference angle
③Example 3you finish the last 2 steps
$ABCD$ is a cyclic quadrilateral. Angle $DAB=98°$. Find angle $x=$ angle $BCD$.
1Name the circle theorem
Opposite angles of a cyclic quadrilateral add up to $180°$
2Write the fact it gives you
angle $DAB$ and angle $BCD$ are OPPOSITE angles of the quadrilateral
3Set up the calculation
$x = 180° - 98°$
4Work out $x$
$x = 82°$
Check · Which pair of angles does the cyclic quadrilateral theorem link?
A angles next to each otherB angles opposite each otherC the two smallest anglesD any angle and the angle at the centre
④Example 4your turn: every step
$TP$ is a tangent touching the circle, centre $O$, at $T$. Angle $TPO=32°$. Find angle $x=$ angle $TOP$.
1Name the circle theorem
A tangent meets a radius at $90°$ at the point of contact
2Write the fact it gives you
angle $OTP=90°$ – mark the right angle at $T$
3Set up the calculation
triangle $OTP$: $32°+90°+x=180°$
4Work out $x$
$x=180°-122°=58°$
Check · Why is the angle at $T$ exactly $90°$?
A because triangle $OTP$ is isoscelesB because angles on a straight line add to $180°$C because $32°+58°=90°$D because a tangent is perpendicular to the radius at the point of contact
Answers · Circle Theorems
Faded examples · Finding angles with circle theorems
① Example 1AB is a diameter, so the angle in a semicircle is a right angle→angle ACB $=90°$ – mark it at C→triangle ACB: $35°+90°+x=180°$→$x=180°-125°=55°$$x = 55°$
Check: A: because AB is a diameter, and the angle in a semicircle is a right angle
② Example 2The angle at the centre is twice the angle at the circumference→angle $AOB$ (at the centre) and angle $x$ (at $C$) stand on the same arc $AB$→$x = 96° \div 2$→$x = 48°$$x = 48°$
Check: C: the centre angle is twice the circumference angle
③ Example 3Opposite angles of a cyclic quadrilateral add up to $180°$→angle $DAB$ and angle $BCD$ are OPPOSITE angles of the quadrilateral→$x = 180° - 98°$→$x = 82°$$x = 82°$
Check: B: angles opposite each other
④ Example 4A tangent meets a radius at $90°$ at the point of contact→angle $OTP=90°$ – mark the right angle at $T$→triangle $OTP$: $32°+90°+x=180°$→$x=180°-122°=58°$$x = 58°$
Check: D: because a tangent is perpendicular to the radius at the point of contact
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