Circles — Important Questions

Answer: (c) $90^{\circ}$.

The angle in a semicircle is a right angle, so $\angle ACB=90^{\circ}$.

Answer: (b) $110^{\circ}$.

Opposite angles of a cyclic quadrilateral are supplementary: $\angle C=180^{\circ}-70^{\circ}=110^{\circ}$.

Answer: (c) $40^{\circ}$.

The angle at the centre is twice the angle at the circumference on the same arc, so the required angle $=\dfrac{80^{\circ}}{2}=40^{\circ}$.

Answer: (b) $105^{\circ}$.

The exterior angle of a cyclic quadrilateral equals the interior opposite angle, so $\angle ADC=\angle CBE=105^{\circ}$.

Answer: (a) Every angle in a segment equals half the central angle standing on the same arc; since they all equal the same half-value, they are equal, so R correctly explains A.

Opposite angles of a cyclic quadrilateral are supplementary.

$\angle BCD=180^{\circ}-\angle DAB=180^{\circ}-92^{\circ}=88^{\circ}.$

$\angle ADC=180^{\circ}-\angle ABC=180^{\circ}-108^{\circ}=72^{\circ}.$

Since $AB$ is a diameter, $\angle ACB=90^{\circ}$ (angle in a semicircle).

In $\triangle ABC$, the angles add to $180^{\circ}$:
$\angle ABC=180^{\circ}-90^{\circ}-35^{\circ}=55^{\circ}.$

For $\angle ACB$ (C on the major arc): the central angle and the inscribed angle stand on the same minor arc $AB$.
$\angle ACB=\tfrac12\,\angle AOB=\tfrac12(100^{\circ})=50^{\circ}.$

For $\angle ADB$ (D on the minor arc): here $ACBD$ form a cyclic quadrilateral, so $\angle ADB$ and $\angle ACB$ are opposite angles.
$\angle ADB=180^{\circ}-\angle ACB=180^{\circ}-50^{\circ}=130^{\circ}.$

(Equivalently, the reflex angle $AOB=260^{\circ}$, and $\angle ADB=\tfrac12(260^{\circ})=130^{\circ}$.)

Given: $AB=CD$; $OM\perp AB$ at $M$ and $ON\perp CD$ at $N$.

To prove: $OM=ON$.

Proof: The perpendicular from the centre to a chord bisects the chord, so
$AM=\tfrac12 AB,\qquad CN=\tfrac12 CD.$
Since $AB=CD$, we get $AM=CN$.

In right triangles $OMA$ and $ONC$:

• $OA=OC$ (radii of the same circle),
• $AM=CN$ (proved above),
• $\angle OMA=\angle ONC=90^{\circ}$.

By the RHS congruence criterion, $\triangle OMA\cong\triangle ONC$, hence $OM=ON$.

Thus equal chords are equidistant from the centre.

Since $ABCD$ is cyclic, opposite angles are supplementary:
$\angle BCD=180^{\circ}-\angle BAD=180^{\circ}-105^{\circ}=75^{\circ}.$

Since $AB\parallel DC$, $\angle BAD$ and $\angle ADC$ are co-interior angles (with transversal $AD$), so they are supplementary:
$\angle ADC=180^{\circ}-105^{\circ}=75^{\circ}.$

Finally, $\angle ABC$ is opposite $\angle ADC$:
$\angle ABC=180^{\circ}-\angle ADC=180^{\circ}-75^{\circ}=105^{\circ}.$

So $\angle ADC=75^{\circ}$, $\angle BCD=75^{\circ}$, $\angle ABC=105^{\circ}$ (it is an isosceles trapezium).

Given: A circle with centre $O$; arc $AB$ subtends $\angle AOB$ at the centre and $\angle ACB$ at a point $C$ on the remaining part of the circle.

To prove: $\angle AOB=2\,\angle ACB$.

Construction: Join $CO$ and produce it to a point $D$.

Proof: In $\triangle OAC$, $OA=OC$ (radii), so $\angle OCA=\angle OAC$. The exterior angle $\angle AOD$ equals the sum of the two interior opposite angles:
$\angle AOD=\angle OCA+\angle OAC=2\,\angle OCA.$

Similarly, in $\triangle OBC$, $OB=OC$, so $\angle OCB=\angle OBC$, giving
$\angle BOD=2\,\angle OCB.$

Adding,
$\angle AOD+\angle BOD=2(\angle OCA+\angle OCB),$
$\angle AOB=2\,\angle ACB.$

Hence the angle at the centre is double the angle at the circumference on the same arc. (The same argument holds when $O$ lies outside the angle, using subtraction instead of addition.)

(i) $\angle ABC$: the reflex angle $AOC=360^{\circ}-130^{\circ}=230^{\circ}$ stands on the major arc; $\angle ABC$ is the inscribed angle on the same major arc as the reflex central angle... more simply, $\angle ABC$ stands on arc $AC$ not containing $B$. Using the central angle $\angle AOC=130^{\circ}$ on the minor arc,
$\angle ABC=\tfrac12\,\angle AOC=\tfrac12(130^{\circ})=65^{\circ}.$

(ii) $\angle ADC$: $ABCD$ is a cyclic quadrilateral, so
$\angle ADC=180^{\circ}-\angle ABC=180^{\circ}-65^{\circ}=115^{\circ}.$

(iii) $\angle BAD$: $BD$ is a diameter, so $\angle BAD=90^{\circ}$ (angle in a semicircle).
(Check with the triangle: in $\triangle ABD$, $\angle ABD=180^{\circ}-90^{\circ}-40^{\circ}=50^{\circ}$.)

(iv) $\angle DBC$: $BD$ is a diameter, so $\angle BCD=90^{\circ}$ (angle in a semicircle). In $\triangle BCD$,
$\angle DBC=180^{\circ}-\angle BCD-\angle BDC=180^{\circ}-90^{\circ}-25^{\circ}=65^{\circ}.$

(i) $\angle QPS$ and $\angle QRS$ are opposite angles of cyclic quadrilateral $PQRS$, so they are supplementary:
$(2x+15)+(3x-10)=180$
$5x+5=180\ \Rightarrow\ 5x=175\ \Rightarrow\ x=35.$

(ii) $\angle QPS=2(35)+15=85^{\circ}$ and $\angle QRS=3(35)-10=95^{\circ}$. (Check: $85^{\circ}+95^{\circ}=180^{\circ}$.)

(iii) $\angle PSR$ is opposite $\angle PQR$, so
$\angle PSR=180^{\circ}-\angle PQR=180^{\circ}-95^{\circ}=85^{\circ}.$

(iv) If $PR$ were a diameter, the angle $\angle PQR$ in the semicircle would be $90^{\circ}$. The surveyor measured $95^{\circ}\neq 90^{\circ}$, so $PR$ is not a diameter; the value is not consistent with $PR$ being a diameter.