(2)

Since PQ is tangent at Q to the circle and OQ is radius,
∴ OQ ? PQ ⇒ ∠OQP = 90°

In ΔPOQ, we have

∠OQP + ∠OPQ + ∠POQ = 180°  [Angle Sum Property of triangle]

⇒ 90° + x + y = 180°
⇒ x + y = 180° – 90° = 90°

(1)

Since TA is tangent at A and OA is radius of the circle,
∴ $OA ⟂ TA \Rightarrow \angle OAT = 90°$

$$\begin{gathered} \therefore In right angled \Delta OAT, we have \\ cos30°=\frac{TA}{OT}=\frac{\sqrt{3}}{2} \\ \frac{TA}{4}=\frac{\sqrt{3}}{2} \\ \Rightarrow TA=2\sqrt{3} \mathrm{cm} \end{gathered}$$

(1)

Let O be the centre of the circle and P be the external point and PA is tangent at A as shown in Fig.

$$\begin{gathered} \therefore OA ⟂ PA \\ Given, OA = 9 cm and OP = 41 cm \\ In right \Delta OAP, we have \\ O{P}^2=O{A}^2+P{A}^2(\mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}) \\ (41{)}^2=(9{)}^2+(PA{)}^2 \\ 1681-81=P{A}^2 \\ 1600=P{A}^2\Rightarrow PA=40 \mathrm{cm} \end{gathered}$$

(1)

Given, OQ = 5 cm

Since PQ is tangent to a circle with centre O, at point Q

$$\begin{gathered} \therefore OQ ⟂ PQ \\ \Rightarrow \angle OQP = 90° \\ In right \Delta OPQ, we have \\ tan30°=\frac{OQ}{PQ} \\ \frac{1}{\sqrt{3}}=\frac{5}{PQ} \\ \Rightarrow PQ=5\sqrt{3} \mathrm{cm} \end{gathered}$$

(2)

Construction: Join OR as shown in

$$\begin{gathered} We have, \Delta OQP \cong \Delta ORP (by SSS congruency rule) \\ \therefore \angle OPQ = \angle OPR = 45° \\ (by CPCT) (\because \angle QPR = 90°) \\ In \Delta POQ, \angle OQP = 90°, \angle OPQ = 45° \\ \angle POQ=180°-(90°+45°)=45° \\ \therefore \angle OPQ = \angle POQ = 45° \\ \therefore \angle OPQ = \angle POQ = 45° \Rightarrow OQ = PQ (Opposite sides of equal angles are equal) \\ \Rightarrow PQ=4 \mathrm{cm} \end{gathered}$$

(4)

Let P be an external point and O be the centre of the circle, such that OP = 2r units where r is radius of the circle as shown in Fig
Let PA be the tangent at point A on circle.

$$\begin{gathered} \therefore OA ⟂ PA and OA = r units \\ Now, in right \Delta OAP, we have \\ A{P}^2+O{A}^2=O{P}^2 \\ A{P}^2+r^2=(2r{)}^2 \\ A{P}^2=4r^2-r^2=3r^2 \\ AP=\sqrt{3} r \mathrm{units} \end{gathered}$$

(1)

In right ΔABC, we have

$$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ A{C}^2=(4{)}^2+(3{)}^2=16+9 \\ A{C}^2=25 \\ AC=5 \mathrm{cm} \\ \mathrm{Now}, \mathrm{AD} = \mathrm{AB} – \mathrm{BD} = (4 – \mathrm{x}) \mathrm{cm} \\ \mathrm{AF}=\mathrm{AD}=\left(4-\mathrm{x}\right) \mathrm{cm}(\mathrm{Tangents} \mathrm{drawn} \mathrm{from} \mathrm{an} \mathrm{external} \mathrm{point} \mathrm{A}) \\ \mathrm{Also}, \\ \mathrm{EC}=\mathrm{BC}\mathrm{\{}–\mathrm{BE}=3-\mathrm{xcm} \\ \mathrm{CF}=\mathrm{EC}=\left(3-\mathrm{x}\right) \mathrm{cm}(\mathrm{Tangents} \mathrm{drawn} \mathrm{from} \mathrm{an} \mathrm{external} \mathrm{point} \mathrm{C}) \\ \mathrm{AC}=\mathrm{AF}+\mathrm{CF}=\left(4-\mathrm{x}\right)+\left(3-\mathrm{x}\right) \\ 5=7-2\mathrm{x} \\ 2\mathrm{x}=7-5=2 \\ \mathrm{x}=1 \mathrm{cm} \end{gathered}$$

(3)

We know that the angle subtended by a chord at the centre is twice the angle subtended by it on the circumference.

$$\begin{gathered} \angle AOC=2\angle ADC \\ 110°=2\angle ADC\Rightarrow \angle ADC=55°\Rightarrow \angle D=55° \end{gathered}$$

$$\begin{gathered} \therefore \angle AOC = 110° \\ \Rightarrow Reflex \angle AOC = 360° – 110° = 250° (As shown in Fig.) \end{gathered}$$

$$\begin{gathered} Also, reflex \angle AOC = 2\angle ABC \\ 250°=2\angle ABC\Rightarrow \angle ABC=125° \end{gathered}$$

$\Rightarrow \angle B = 125° \Rightarrow Statements \left(i\right) and \left(iv\right) are correct.$

(4)

In a cyclic quadrilateral, the sum of opposite angles is 180°. Only rectangle satisfies the condition of cyclic quadrilateral.