iGCSE Mathematics (0580) - C4.7 Circle theorems- Exam Style Questions Paper 1- New Syllabus

Question

Points $A$, $B$ and $C$ lie on the circle, centre $O$.

Work out angle $BCA$.

Most-appropriate topic codes (Cambridge IGCSE Mathematics 0580):

• C4.7 Circle theorems

▶️ Answer/Explanation

Since the line $AC$ passes directly through the circle’s centre $O$, it is a diameter. By applying the circle theorem rule that states the angle subtended by a semicircle is always a right angle, we know angle $ABC = 90^\circ$. Because the interior angles of any triangle add up to $180^\circ$, we compute angle $BCA = 180^\circ – 90^\circ – 39^\circ = 51^\circ$.
✅ Answer: $51^\circ$

Question

$B$ is a point on the circle, centre $O$.
$ABC$ is a tangent to the circle at $B$ and $\angle OAB = 36°$. [NOT TO SCALE]

Work out the size of angle $AOB$.

Most-appropriate topic codes (Cambridge IGCSE Mathematics 0580):

• TOPIC C4.7 Circle theorems: The angle between a tangent and a radius = 90°; use this to calculate unknown angles in circle problems (Core)

▶️ Answer/Explanation

Since $ABC$ is a tangent to the circle at $B$, and $OB$ is a radius, the tangent-radius theorem tells us $\angle OBA = 90°$.
In triangle $OAB$, the angles must sum to $180°$: $\angle AOB = 180° – 90° – 36° = 54°$.
✅ Answer: $\angle AOB = 54°$

Question

The diagram shows a circle, centre $O$.
$P$ and $S$ are points on the circle.
$POR$ is a straight line.
$QRST$ is a tangent to the circle at $S$. Angle $OPS = 25^{\circ}$.

(a) Find the value of $x$. Give a geometrical reason for your answer.

(b) Find the value of $y$. [Note: $y$ represents angle $ORS$ from the context].

Most-appropriate topic codes (Cambridge IGCSE Mathematics 0580):

• C4.7 Circle theorems

▶️ Answer/Explanation

(a) Because segments $OP$ and $OS$ are both radii extending from the center of the same circle, the triangle $\triangle OPS$ is isosceles. Since base angles of an isosceles triangle are equal, angle $x$ (which is angle $OSP$) must equal angle $OPS = 25^{\circ}$.

(b) A key circle theorem states that a tangent meets a radius at an exact $90^{\circ}$ angle, meaning $\angle OSR = 90^{\circ}$. In the large triangle $\triangle PSR$ or by looking at the outer angle components:
Using the straight line line $POR$: $\angle POS = 180^{\circ} – (25^{\circ} + 25^{\circ}) = 130^{\circ}$. Therefore, the adjacent angle $\angle SOR = 180^{\circ} – 130^{\circ} = 50^{\circ}$.
Finally, inside right-angled triangle $\triangle OSR$, angles add up to $180^{\circ}$: $$y = 180^{\circ} – (90^{\circ} + 50^{\circ}) = 140^{\circ}\text{ (exterior angle logic gives } 90 + 25 + 25 = 140^{\circ}\text{)}$$

Answers:
(a) $x =$ 25 because base angles of an isosceles triangle are equal
(b) $y =$ 140

iGCSE Mathematics (0580) - C4.7 Circle theorems- Exam Style Questions Paper 1- New Syllabus

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