17. Tangents and Intersecting Chords Previous Year Questions

(a) In right ΔOBX (∠OXB = 90°): ∠BOX = 90° − 32° = 58°. [1 Mark]
(b) By Alternate Segment Theorem, angle between tangent AX and chord XZ equals angle in alternate segment: ∠XYZ = ∠AXZ = 66°. [1 Mark]
(c) In ΔXYZ, ZY = XY ⇒ ∠XZY = ∠ZXY = (180° − 66°)/2 = 57°. Hence ∠ZYX = 66°. [1 Mark]
(d) Since OX ⊥ AB, ∠AXO = 90°. ∠OXY = ∠AXZ − ∠AXO = 66° − 90° or in ΔOXY, ∠OXY = 33°. [1 Mark]

(a) In ΔPTR and ΔPST: ∠P = ∠P (Common), ∠PTR = ∠PST (Angle in alternate segment theorem). By AA criterion, ΔPTR ∼ ΔPST. [1 Mark]
(b) Proportional sides: PT/PS = PR/PT ⇒ PT² = PR × PS. [1 Mark]
(c) PT² = 4 × 16 = 64 ⇒ PT = 8 cm. [1 Mark]

Correct Option: (d) 50°
In ΔPST, PS = PT ⇒ ∠PTS = (180°−80°)/2 = 50°.
Since SQ ∥ PT, alternate interior angle ∠QST = ∠PTS = 50°. [1 Mark]

∠AOC = 2 × 65° = 130°.
(a) ∠ABC = ½∠AOC = 65°. [1 Mark]
(b) ΔABC is isosceles ⇒ ∠ACB = 65° ⇒ ∠BAC = 180° − 130° = 50°. [1 Mark]
(c) Alternate segment theorem: ∠BCQ = ∠BAC = 50°. [1 Mark]

Radius = 9 cm ⇒ Diameter AB = 18 cm. PB = PA − AB = 24 − 18 = 6 cm. [1 Mark]
By Tangent-Secant Theorem: PQ² = PA × PB = 24 × 6 = 144. [1 Mark]
PQ = √144 = 12 cm. [1 Mark]

Step 1 By the Alternate Segment Theorem, the angle between the tangent $ST$ and chord $SP$ equals the angle subtended by $SP$ in the alternate segment:
$\angle PQS = \angle PST = 30^\circ$.
Step 2 In $\triangle PQS$, the sum of interior angles is $180^\circ$:
$\angle PSQ = 180^\circ - (\angle SPQ + \angle PQS) = 180^\circ - (60^\circ + 30^\circ) = 180^\circ - 90^\circ = 90^\circ$.
Correct Option: (d) [1 Mark]

Since segment $AB$ passes through the centre $O$, $AB$ is a diameter.

(a) $\angle BDA$ Angle in a semicircle is a right angle: $\angle BDA = 90^\circ$.1 Mark

(b) $\angle BAD$ In right-angled $\triangle ABD$, the acute angles are complementary:
$\angle BAD = 90^\circ - \angle ABD = 90^\circ - 26^\circ = \mathbf{64^\circ}$.1 Mark

(c) $\angle CAD$ By the Alternate Segment Theorem, the angle between tangent $AC$ and chord $AD$ is equal to the angle subtended by chord $AD$ in the alternate segment:
$\angle CAD = \angle ABD = \mathbf{26^\circ}$.
(Alternatively: radius $OA \perp CE \implies \angle CAO = 90^\circ$. Thus $\angle CAD = 90^\circ - 64^\circ = 26^\circ$).1 Mark

(d) $\angle ODB$ In $\triangle OBD$, $OB = OD$ (radii of the same circle).
Therefore, $\triangle OBD$ is isosceles with $\angle ODB = \angle OBD = \mathbf{26^\circ}$.1 Mark

(a) Length of PT By the Tangent-Secant Theorem, $PT^2 = PA \times PB$.
Here $PA = AB + BP = 9 + 16 = 25\text{ cm}$, and $PB = 16\text{ cm}$.
$PT^2 = 25 \times 16 = 400 \implies PT = \sqrt{400} = \mathbf{20\text{ cm}}$.1 Mark

(b) $\angle BAT$ By the Alternate Segment Theorem, the angle between tangent $PT$ and chord $BT$ equals the angle subtended by $BT$ in the alternate segment:
$\angle BAT = \angle PTB = \mathbf{50^\circ}$.1 Mark

(c) $\angle BOT$ The angle subtended by arc $BT$ at the centre is twice the angle subtended by it at any point on the remaining part of the circle:
$\angle BOT = 2 \times \angle BAT = 2 \times 50^\circ = \mathbf{100^\circ}$.1 Mark

(d) $\angle ABT$ In $\triangle OBT$, $OB = OT$ (radii). Thus $\triangle OBT$ is isosceles:
$\angle OBT = \frac{180^\circ - \angle BOT}{2} = \frac{180^\circ - 100^\circ}{2} = 40^\circ$.
Radius $OB$ divides angle $\angle ABT$ into $\angle OBA$ and $\angle OBT$:
$\angle ABT = \angle OBA + \angle OBT = 45^\circ + 40^\circ = \mathbf{85^\circ}$.1 Mark

Tangent radius theorem: $OA \perp TP \implies \angle OAP = 90^\circ$ and $OC \perp TQ \implies \angle OCQ = 90^\circ$.
(i) In $\triangle OAB$: $OA = OB$ (radii).
$\angle OAB = 90^\circ - \angle BAP = 90^\circ - 60^\circ = 30^\circ \implies \mathbf{\angle OBA = 30^\circ}$.
In $\triangle OBC$: $OC = OB$ (radii).
$\angle OCB = 90^\circ - \angle BCQ = 90^\circ - 55^\circ = 35^\circ \implies \mathbf{\angle OBC = 35^\circ}$. [1.5 Marks]
(ii) Total angle $\angle ABC = \angle OBA + \angle OBC = 30^\circ + 35^\circ = 65^\circ$.
Angle at centre is double the angle at the circumference:
$\mathbf{\angle AOC = 2 \times \angle ABC = 2 \times 65^\circ = 130^\circ}$. [1.5 Marks]
(iii) In tangent quadrilateral $OATC$: $\angle OAT = \angle OCT = 90^\circ$.
$\mathbf{\angle ATC = 180^\circ - \angle AOC = 180^\circ - 130^\circ = 50^\circ}$. [1 Mark]

Step 1 Find $\angle B$ using properties of circle and tangent:
$AC$ is tangent at $A$ and $OA$ is radius $\implies OA \perp AC \implies \angle OAC = 90^\circ$.
In $\Delta ABD$, angle in semicircle / tangent property gives $\angle A = 90^\circ$.
In $\Delta ABD$: $\angle B = 180^\circ - (\angle A + \angle ADB) = 180^\circ - (90^\circ + 55^\circ) = 180^\circ - 145^\circ = 35^\circ$.
Since $OB = OE$ (radii of circle), in $\Delta OBE$, $\angle OEB = \angle B = 35^\circ$. [1 Mark]
Step 2 Find $x$ using $\Delta EDC$:
$\angle DEC = \angle OEB = 35^\circ$ (vertically opposite angles).
Since $\angle ADE$ and $\angle EDC$ form a linear pair on line $AD$:
$\angle EDC = 180^\circ - 55^\circ = 125^\circ$.
In $\Delta EDC$: the sum of angles is $180^\circ$:
$x + \angle DEC + \angle EDC = 180^\circ \implies x + 35^\circ + 125^\circ = 180^\circ \implies x + 160^\circ = 180^\circ \implies \mathbf{x = 20^\circ}$. [1 Mark]
Step 3 Find $y$ using right triangle $\Delta OAC$:
In $\Delta OAC$, $\angle OAC = 90^\circ$ and $\angle ACO = x = 20^\circ$.
$y + \angle OAC + x = 180^\circ \implies y + 90^\circ + 20^\circ = 180^\circ \implies y + 110^\circ = 180^\circ \implies \mathbf{y = 70^\circ}$.
Answer: $x = 20^\circ$ and $y = 70^\circ$. [1 Mark]

Step 1 (a) Prove $\Delta PTR \sim \Delta QTP$:
In $\Delta PTR$ and $\Delta QTP$:
$\angle T = \angle T$ (Common angle)
$\angle TPR = \angle TQP$ (Angles in alternate segment by Alternate Segment Theorem)
$\therefore \mathbf{\Delta PTR \sim \Delta QTP}$ by **AA Similarity Criterion**. [2 Marks]
Step 2 (b) Determine length of $PT$ using ratio of corresponding sides:
$\frac{PT}{QT} = \frac{RT}{PT} \implies PT^2 = QT \times RT$.
Given $QR = 9\text{ cm}$ and $RT = 16\text{ cm} \implies QT = QR + RT = 9 + 16 = 25\text{ cm}$.
$PT^2 = 25 \times 16 = 400 \implies PT = \sqrt{400} = \mathbf{20\text{ cm}}$. [2 Marks]

(i) Prove BD is Diameter
Since $AB$ bisects $\angle CAQ$: $\angle CAB = \angle BAQ = 30^\circ \implies \angle CAQ = 60^\circ$.
Since $PAQ$ is a straight tangent line: $\angle PAC = 180^\circ - 60^\circ = 120^\circ$.
$AD$ bisects $\angle PAC \implies \angle DAC = \frac{120^\circ}{2} = 60^\circ$.
$$\angle DAB = \angle DAC + \angle CAB = 60^\circ + 30^\circ = 90^\circ$$ Since $\angle DAB = 90^\circ$ is an inscribed angle in the circle, chord $BD$ must subtend a right angle at the circumference.
$\therefore \mathbf{BD\text{ is a diameter of the circle}}$. [1.5 Marks]
(ii) Prove ABC is Isosceles
By Alternate Segment Theorem: Angle between tangent $AQ$ and chord $AB$ equals the angle in the alternate segment:
$$\angle ACB = \angle BAQ = 30^\circ$$ In $\triangle ABC$, $\angle CAB = 30^\circ$ and $\angle ACB = 30^\circ$.
Since two angles are equal, opposite sides $AB = BC$.
$\therefore \mathbf{\triangle ABC\text{ is an isosceles triangle}}$. [1.5 Marks]
Hence Proved.