Circle Theorems practice questions — Honors Math

Since $P$ lies on the minor arc $AB$, the angle $\mathrm{\angle }APB$ 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: $\text{reflex }\mathrm{\angle }AOB=2\times {118}^{\circ }={236}^{\circ }$.

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

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

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

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

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

(Equivalently, the exterior angle $\mathrm{\angle }DCE$ 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:

$\mathrm{\angle }DBA=\mathrm{\angle }DAB={58}^{\circ }$.

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

$$\mathrm{\angle }ADB={64}^{\circ }$$

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

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

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

$\mathrm{\angle }ACB={90}^{\circ }$.

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

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

$$\mathrm{\angle }BAC={36}^{\circ }$$

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

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

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 $\mathrm{\angle }ABD$ and $\mathrm{\angle }ACD$ both stand on the chord $AD$ from that same segment, so they are equal: $$\mathrm{\angle }ACD={56}^{\circ }\mathrm{.}$$

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

$\mathrm{\angle }DAB$ and $\mathrm{\angle }BCD$ 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 $\mathrm{\angle }BCD$, so substitute into that expression only:

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

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

$$\mathrm{\angle }BCD={70}^{\circ }$$

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 $\mathrm{\angle }SAB$: $$\mathrm{\angle }ACB=\mathrm{\angle }SAB={68}^{\circ }\mathrm{.}$$ Since $BC\parallel ST$ and $AB$ is a transversal, $\mathrm{\angle }ABC$ and $\mathrm{\angle }SAB$ are alternate angles, so $\mathrm{\angle }ABC={68}^{\circ }$. In triangle $ABC$, $\mathrm{\angle }BAC={180}^{\circ }-{68}^{\circ }-{68}^{\circ }$. $$\mathrm{\angle }BAC={44}^{\circ }$$

Because $AB$ is a diameter, the point $C$ lies on a semicircle, so the angle it subtends there is a right angle: $\mathrm{\angle }ACB={90}^{\circ }$.

The interior angles of triangle $ABC$ sum to ${180}^{\circ }$: $\mathrm{\angle }BAC={180}^{\circ }-\mathrm{\angle }ACB-\mathrm{\angle }ABC={180}^{\circ }-{90}^{\circ }-{27}^{\circ }$. $$\mathrm{\angle }BAC={63}^{\circ }\mathrm{.}$$

Going round the circle $A\to B\to C\to D$, the vertex opposite $C$ is $A$, and the vertex opposite $D$ is $B$.

Opposite angles of a cyclic quadrilateral are supplementary, so $\mathrm{\angle }DAB+\mathrm{\angle }BCD={180}^{\circ }$:

$\mathrm{\angle }BCD={180}^{\circ }-{68}^{\circ }={112}^{\circ }$.

The angle $\mathrm{\angle }ABC={95}^{\circ }$ belongs to the other pair, and it fixes the fourth angle instead: $\mathrm{\angle }ADC={180}^{\circ }-{95}^{\circ }={85}^{\circ }$. As a check, ${68}^{\circ }+{95}^{\circ }+{112}^{\circ }+{85}^{\circ }={360}^{\circ }$.

$$\mathrm{\angle }BCD={112}^{\circ }$$

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

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