Circle Theorems practice questions — GCSE Maths

In triangle $PAC$ the angles at $P$ and $C$ are $34^\circ$ and $48^\circ$, so $$\angle PAC = 180^\circ - 34^\circ - 48^\circ = 98^\circ.$$ The tangent $PA$ and the chord $AB$ meet at $A$; the alternate segment for that angle contains $C$, so $\angle PAB = \angle ACB = 48^\circ$. Because $B$ lies between $P$ and $C$, the ray $AB$ lies inside $\angle PAC$, hence $\angle BAC = \angle PAC - \angle PAB = 98^\circ - 48^\circ$. $$\angle BAC = 50^\circ$$

$C$ and $D$ both lie on the arc $AB$ on the far side of the chord $AB$ from $S$, so that arc is the alternate segment for $\angle SAB$: $$\angle ADB = \angle SAB = 41^\circ.$$ The cyclic order $A$, $B$, $C$, $D$ puts the ray $DB$ inside $\angle ADC$, so $\angle ADC = \angle ADB + \angle BDC = 41^\circ + 27^\circ = 68^\circ$. $ABCD$ is a cyclic quadrilateral, so $\angle ABC$ and $\angle ADC$ are opposite angles and sum to $180^\circ$: $\angle ABC = 180^\circ - 68^\circ$. $$\angle ABC = 112^\circ$$

$\angle DAB$ and $\angle BCD$ sit at opposite vertices of the cyclic quadrilateral, so they are supplementary:

$(3x - 10) + (x + 30) = 180$

$4x + 20 = 180$

$4x = 160$, so $x = 40$.

The question asks for $\angle BCD$, so substitute into that expression only:

$\angle BCD = (40 + 30)^\circ = 70^\circ$.

Check: $\angle DAB = (3 \cdot 40 - 10)^\circ = 110^\circ$, and $110^\circ + 70^\circ = 180^\circ$.

$$\angle BCD = 70^\circ$$

$P$ and $C$ lie on opposite sides of the chord $AB$, so the segment containing $C$ is the alternate segment for the tangent-chord angle $\angle PAB$: $$\angle ACB = \angle PAB = 58^\circ.$$ In triangle $ABC$ the three angles are $\angle BAC$, $\angle ABC = 47^\circ$ and $\angle ACB = 58^\circ$, so $\angle BAC = 180^\circ - 47^\circ - 58^\circ$. $$\angle BAC = 75^\circ$$

The interior angles of triangle $ABX$ sum to $180^\circ$: $\angle ABX = 180^\circ - 28^\circ - 96^\circ = 56^\circ$.

The point $X$ lies on the chord $BD$, so the ray $BX$ is the ray $BD$ and therefore $\angle ABD = \angle ABX = 56^\circ$.

Because the chords $AC$ and $BD$ cross inside the circle, $B$ and $C$ lie on the same arc determined by $A$ and $D$. The angles $\angle ABD$ and $\angle ACD$ both stand on the chord $AD$ from that same segment, so they are equal: $$\angle ACD = 56^\circ.$$

$B$, $C$ and $E$ are collinear, so $\angle BCD$ and $\angle DCE$ are angles on a straight line:

$\angle BCD = 180^\circ - 58^\circ = 122^\circ$.

$\angle DAB$ is opposite $\angle BCD$ in the cyclic quadrilateral $ABCD$, so the two are supplementary:

$\angle DAB = 180^\circ - 122^\circ = 58^\circ$.

(Equivalently, the exterior angle $\angle DCE$ equals the interior angle at the opposite vertex $A$.)

In triangle $ABD$, the equal sides are $DA$ and $DB$, so the angles opposite them — the base angles at $B$ and at $A$ — are equal:

$\angle DBA = \angle DAB = 58^\circ$.

$\angle ADB = 180^\circ - 58^\circ - 58^\circ = 64^\circ$.

$$\angle ADB = 64^\circ$$

$\angle ADC$ and $\angle ABC$ are opposite angles of the cyclic quadrilateral $ABCD$, so they are supplementary:

$\angle ABC = 180^\circ - 126^\circ = 54^\circ$.

$C$ lies on the circle and $AB$ is a diameter, so $AB$ subtends a right angle at $C$:

$\angle ACB = 90^\circ$.

In triangle $ABC$ the three angles sum to $180^\circ$:

$\angle BAC = 180^\circ - 90^\circ - 54^\circ = 36^\circ$.

$$\angle BAC = 36^\circ$$

Since $P$ lies on the minor arc $AB$, the angle $\angle APB$ stands on the major arc $AB$. The angle at the centre standing on that same major arc is the reflex angle at $O$, and it is twice the angle at the circumference: $\text{reflex } \angle AOB = 2 \times 118^\circ = 236^\circ$.

The reflex and non-reflex angles at $O$ together make a complete turn: $\angle AOB = 360^\circ - 236^\circ$. $$\angle AOB = 124^\circ.$$

The ray $AC$ lies between $AB$ and $AT$, so $C$ is on the opposite side of the chord $AB$ from $S$; the segment holding $C$ is therefore the alternate segment for $\angle SAB$: $$\angle ACB = \angle SAB = 68^\circ.$$ Since $BC \parallel ST$ and $AB$ is a transversal, $\angle ABC$ and $\angle SAB$ are alternate angles, so $\angle ABC = 68^\circ$. In triangle $ABC$, $\angle BAC = 180^\circ - 68^\circ - 68^\circ$. $$\angle BAC = 44^\circ$$

A tangent is perpendicular to the radius drawn to its point of contact, so $\angle OAP = \angle OBP = 90^\circ$.

The four points $O$, $A$, $P$, $B$ form a quadrilateral, whose interior angles sum to $360^\circ$: $\angle AOB = 360^\circ - \angle OAP - \angle OBP - \angle APB = 360^\circ - 90^\circ - 90^\circ - 48^\circ$. $$\angle AOB = 132^\circ.$$