We are given a circle with centre O and a tangent XY to the circle at a point P. We need to prove that OP is perpendicular to XY.

Take a point Q on XY other than P and join OQ see  Fig.The point Q must lie outside the circle. (Why? Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle). Therefore, OQ is longer than the radius OP of the circle. That is, OQ > OP.
Since this happens for every point on the line XY except the point P, OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY.

\( \frac{3}{2}\) 

Given, \(\sin { \alpha =\frac { 1 }{ 2 } } \)
then \(3\sin { \alpha } -4\sin ^{ 3 }{ \alpha } =3\times \frac { 1 }{ 2 } -4\times { \left( \frac { 1 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 3 }{ 2 } -\frac { 4 }{ 8 } =1\)  

Given, umbrella to be a flat circle. So, the central angle of an umbrella is 360°.
Since, umbrella has 8 ribs.
\(\therefore\) Angle between two ribs.
\(=\frac{360^{\circ}}{8}=45^{\circ}\)
Area between two ribs = Area of one sector of the umbrella
\(=\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(45)^2[\because r=45, \text { given }]\)
\(\begin{aligned} & =\frac{22}{7 \times 8}(45)^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22275}{28} \mathrm{~cm}^2 \end{aligned}\)

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.
In 5 minutes, minute hand will rotate = 360^@/60xx5 = 30^@
Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.
Area of sector of angle θ = \(\frac{\theta}{360^{\circ}} \times \pi r^{2}\)
Area of sector of 30° \(=\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14\)
\(\begin{array}{l} =\frac{22}{12} \times 2 \times 14 \\ =\frac{11 \times 14}{3} \end{array}\)
=154/3 cm²
Therefore, the area swept by the minute hand in 5 minutes is 154/3 cm²

We know that area of sector of a circle =\(\frac{\theta}{360^{\circ}} \times \pi r^2\)
Given, radius of circle, r = 6 cm
and angle of sector, \(\theta\)= 60°
\(\therefore\) Area of sector of a circle \(=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(6)^2=\frac{132}{7} \mathrm{~cm}^2\)

Let AB be a diameter of a given circle and LM and PQ be the tangent lines drawn to the circle at points A and B, respectively.

To prove LM || PQ
Proof We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(\therefore\) OA \(\perp\) PQ and OB \(\perp\) LM
\(\Rightarrow\) AB \(\perp\) PQ
and AB \(\perp\) LM
\(\Rightarrow\) \(\angle\)PAB = 90°
and  \(\angle\)ABM = 90°
\(\Rightarrow\) \(\angle\)PAB = \(\angle\)ABM
[each = 90°]
But these are alternate angles.
\(\therefore\) PQ || LM
Hence, the tangents drawn at the ends of a diameter of a circle are parallel.
Hence proved.

In right angled \(\triangle PQR,\)
PQ = 12 cm, PR = 13 cm   [given]
Then, PQ² + QR² = PR²   [by using Pythagoras theorem]
\(\Rightarrow\)  (12)² + QR² = (13)² 
\(\Rightarrow\)  144 + QR² = 169
\(\Rightarrow\)  QR² = 169 - 144 = 25
\(\Rightarrow\)  QR = 5 [taking positive square root since, side cannot be negative]
Now, \(tanP=\frac { P }{ B } =\frac { QR }{ PQ } =\frac { 5 }{ 12 } \)
\(and\quad cot \quad R=\frac { B }{ P } =\frac { QR }{ PQ } =\frac { 5 }{ 12 } \)
\(\therefore \quad tanP-cosR=\frac { 5 }{ 12 } - \frac { 5 }{ 12 } =0\) 

Area of the segment AYB = Area of sector OAYB - Area of \(\Delta\)OAB                  (1)
Now, area of the sector OAYB = \(=\frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^2=462 \mathrm{~cm}^2\) (2)
For finding the area of \(\Delta\)OAB, draw OM \(\perp\) AB as shown in Figure.
Note that OA = OB. Therefore, by RHS congruence, \(\Delta\)AMO \(\cong\) BMO.
So, M is the mid-point of AB and \(\angle\)AOM = \(\angle\)BOM = \(\frac{1}{2} \times 120^{\circ}=60^{\circ}\)
Let  OM = x cm
So, from \(\Delta\)OMA, \(\frac{\mathrm{OM}}{\mathrm{OA}}=\cos 60^{\circ}\)

or, \(\begin{aligned} \frac{x}{21} & =\frac{1}{2} \quad\left(\cos 60^{\circ}=\frac{1}{2}\right) \\ \end{aligned}\)
or, \(\begin{aligned} x & =\frac{21}{2} \\ \end{aligned}\)
So, \(\begin{aligned} \mathrm{OM} & =\frac{21}{2} \mathrm{~cm} \\ \end{aligned}\)
Also,\(\begin{aligned} \frac{\mathrm{AM}}{\mathrm{OA}} & =\sin 60^{\circ}=\frac{\sqrt{3}}{2} \\ \end{aligned}\)
So, \(\begin{aligned} \mathrm{AM} & =\frac{21 \sqrt{3}}{2} \mathrm{~cm} \end{aligned}\)
Therefore, \(\mathrm{AB}=2 \mathrm{AM}=\frac{2 \times 21 \sqrt{3}}{2} \mathrm{~cm}=21 \sqrt{3} \mathrm{~cm}\)
So, area of \(\begin{aligned} \Delta \mathrm{OAB} & =\frac{1}{2} \mathrm{AB} \times \mathrm{OM}=\frac{1}{2} \times 21 \sqrt{3} \times \frac{21}{2} \mathrm{~cm}^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{441}{4} \sqrt{3} \mathrm{~cm}^2 \end{aligned}\)                                                                              (3)
Therefore, are of the segment \(\mathrm{AYB}=\left(462-\frac{441}{4} \sqrt{3}\right) \mathrm{cm}^2\) [From (1), (2) and (3)]
\(=\frac{21}{4}(88-21 \sqrt{3}) \mathrm{cm}^2\)

Let AD = 7 m be the height of the building and BC = h m be the height of the cable tower. From the top of the building D, the angles of elevation and depression are \(\angle\)CDE = 60^o and \(\angle\)EDB = 45^o    
From the point D, draw a line DE || AB.
Then, \(\angle\)EDB = \(\angle\)ABD = 45^o  [alternate angles]

Also, let AB = DE = x m be the distance between building and tower.
In right angled \(\Delta\)BAD,
\(\begin{array}{rlrl} \tan 45^{\circ} & =\frac{P}{B}=\frac{A D}{A B} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad 1 & =\frac{7}{x} & {\left[\because \tan 45^{\circ}=1\right]} \end{array}\)
\(\Rightarrow\) x = 7 m   ....(i)
and in right angled \(\Delta\)CED,
\(\begin{aligned} & \tan 60^{\circ}=\frac{C E}{D E}=\frac{C B-B E}{A B}[\because C E=C B-B E] \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \sqrt{3}=\frac{b-7}{x} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right\} \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad b-7=x \sqrt{3} \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad h=x \sqrt{3}+7 \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad h=7 \sqrt{3}+7 \quad \text { [from Eq. (i)] } \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad h=7(\sqrt{3}+1) \mathrm{m} \\ \end{aligned}\)
Hence, the height of the tower is \(7(\sqrt{3}+1) \mathrm{m} .\)

Let BC be the building, AB be the transmission tower and D be the point on the ground from where the angles of elevations are to be measured.

\(\begin{array}{rlrl} \text { In } \triangle B C D, & \tan 45^{\circ} =\frac{B C}{C D} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow 1 =\frac{20}{C D} \Rightarrow C D=20 \mathrm{~m} \end{array}\)
\(\begin{array}{llrl} \text { In } \triangle A C D, & \tan 60^{\circ} =\frac{A C}{C D} \end{array}\)
\(\begin{array}{llrl} \Rightarrow \sqrt{3} =\frac{A B+B C}{C D} \\ \end{array}\)
\(\begin{array}{llrl} \Rightarrow \sqrt{3} =\frac{A B+20}{20} \\ \end{array}\)
\(\Rightarrow A B =20 \sqrt{3}-20=20(\sqrt{3}-1) \mathrm{m}\)
Thus, the height of the tower is \(20(\sqrt{3}-1) \mathrm{m}\).

Given XY and X'Y' are two parallel tangents. Another tangent AB touches the circle at C and intersect XY at A and X'Y'at B.
To prove \(\angle\)AOB = 90°
Proof We know that, tangents drawn from an external point to a circle are equal in length.
\(\therefore\)   AP = AC   [\(\because\) A is an external point] ...(i)
Thus, in \(\Delta\)APO and \(\Delta\)ACO, AP = AC   [from Eq. (i)]
AO = AO  [common sides]
OP = OC [radii of circle]
\(\Delta\)APO \(\cong\)\(\Delta\)ACO  [by SSS congruence rule]
Then, \(\angle\)OAP = \(\angle\)OAC    [by CPCT]        ...(ii)
\(\Rightarrow\) \(\angle\)PAC = 2 \(\angle\)CAO                             ....(iii)
Similarly, we can prove that \(\angle\)CBO = \(\angle\)OBQ
\(\Rightarrow\)   \(\angle\)CBQ = 2 \(\angle\)CBO                           ...(iv)
since, XY || X'Y'      [given]
\(\therefore\) \(\angle\)PAC +  \(\angle\)QBC = 180°
[\(\because\) sum of interior angles on the same side of transversal is 180°]
\(\Rightarrow\) 2 \(\angle\)CAO + 2 \(\angle\)CBO = 180°    [from Eqs. (iii) and (iv)]
\(\Rightarrow\)   \(\angle\)CAO + \(\angle\)CBO = 90°                    ...(v)
Now, in \(\Delta\)AOB, \(\angle\)CAO + \(\angle\) CBO + \(\angle\)AOB = 180°
[by angle sum property of triangle]
\(\Rightarrow\) \(\angle\)CAO + \(\angle\)CBO = 180° - \(\angle\)AOB    ...(vi)
From Eqs. (v) and (vi), we get
180° - \(\angle\)AOB = 90° \(\Rightarrow\) \(\angle\)AOB = 90° Hence proved.

Given A quadrilateral ABCD, circumscribing a circde.
To prove AB + CD = AD + BC
Proof We know that the lengths of tangents drawn from an external point to a circde are equal.
\(\therefore\)   AP = AS               ...(i)
[\(\because\) both are tangents to a circle from point A]
Similarly, BP = BQ,   ...(ii)
CR = CQ                  ....(iii)
and DR = DS           .....(iv)
On adding Eqs. (i), (ii), (ii) and (iv), we get
(AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS)
\(\Rightarrow\)   AB + CD = (AS + DS) + (BQ + CQ)
\(\Rightarrow\)  AB + CD = AD + BC
Hence Proved.

sin 60° cos 30° + sin 30° cos 60°
\(=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\)
\(\left[\because \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2} \text { and } \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\right]\)
\(=\frac{3}{4}+\frac{1}{4}=\frac{3+1}{4}=\frac{4}{4}=1\)

Given PQ = 3 cm and PR = 6 cm.
Therefore,  \(\begin{aligned} \frac{\mathrm{PQ}}{\mathrm{PR}} & =\sin \mathrm{R} \\ \end{aligned}\)
or  \(​​​​\begin{aligned} \sin \mathrm{R} & =\frac{3}{6}=\frac{1}{2} \\ \end{aligned}\)
So, \(\begin{aligned} \angle \mathrm{PRQ} & =30^{\circ} \\ \end{aligned}\)
and therefore, \(\begin{aligned} \angle \mathrm{QPR} & =60^{\circ} \end{aligned}\)
You may note that if one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be determined.

Let us first draw a right \(\Delta\)ABC

Now, we know that tan \(\mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{4}{3}\)
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Now, by using the Pythagoras Theorem, we have
AC² = AB² + BC² = (4k)² + (3k)² = 25k²
So, AC = 5k
Now, we can write all the trigonometric ratios using their definitions.
\(\begin{aligned} & \sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{4 k}{5 k}=\frac{4}{5} \\ \end{aligned}\)
\(\begin{aligned} & \cos \mathrm{A}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3 k}{5 k}=\frac{3}{5} \end{aligned}\)
Therefore, cot \(\mathrm{A}=\frac{1}{\tan \mathrm{A}}=\frac{3}{4}, \operatorname{cosec} \mathrm{A}=\frac{1}{\sin \mathrm{A}}=\frac{5}{4} \text { and } \sec \mathrm{A}=\frac{1}{\cos \mathrm{A}}=\frac{5}{3} .\)

since, sin (A - B) = \(\frac{1}{2}\), therefore, A - B = 30°              (1)
Also, since cos (A + B) = \(\frac{1}{2}\), therefore, A + B = 60°   (2)
Solving (1) and (2), we get : A = 45° and B = 15°.

92.4 cm²

81.75 cm²

We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P see fig. We are required to prove that PQ = PR.

For this, we join OP, OQ and OR. Then \(\angle\)OQP and \(\angle\)ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP,
OQ = OR (Radii of the same circle)
OP = OP (Common)
Therefore,   \(\Delta\)OQP \(\cong\)\(\Delta\) ORP   (RHS)
This gives  PQ = PR    (CPCT)

Given, diameter of circle, d = 35 mm
\(\therefore\) Circumference of circle = \(\pi\)d      [\(\because\) d = 2r]
\(=\frac{22}{7} \times 35=110 \mathrm{~mm}^2\)
Now, length of 5 diameters = 5 \(\times\) 35 = 175 mm
(i) Total length of the silver wire = \(\pi\)d + 5d
= 110 + 175 = 285 mm²
(ii) Here, we see that total circle is divided into 10 sectors.
\(\therefore\) Angle of each sector = \(\frac{360^{\circ}}{10}=36^{\circ}\)
Then, area of each sector ofthe brooch = \(=\frac{\theta}{360^{\circ}} \times \pi r^2\)
\(\begin{aligned} & =\frac{36^{\circ}}{360^{\circ}} \times \frac{22}{7}\left(\frac{35}{2}\right)^2 \quad\left[\because r=\frac{d}{2}=\frac{35}{2} \mathrm{~mm}\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{10} \times \frac{22}{1} \times \frac{5}{2} \times \frac{35}{2}=\frac{11 \times 35}{2 \times 2}=\frac{385}{4} \mathrm{~mm}^2 \end{aligned}\)

If Assertion is incorrect but Reason is correct.

(d) If Assertion is incorrect but Reason is correct.

(i) (c) In quadrilateral PQOR, we have

\(\begin{array}{r} \angle Q P R+\angle P R O+\angle P Q O+\angle R O Q=360^{\circ} \\ \end{array}\)
\(\Rightarrow \begin{array}{r} 30^{\circ}+90^{\circ}+90^{\circ}+\angle R O Q=360^{\circ} \end{array}\)
[\(\because\) radius is always perpendicular to the tangent at point of contact]
\(\Rightarrow \quad \angle R O Q=360^{\circ}-210^{\circ}=150^{\circ} \)
\((ii) (a) In \quad \triangle Q O R, O Q=O R\)     [radii]
\(\begin{array}{ll} \therefore & \angle O R Q=\angle O Q R \\ \end{array}\)
\(\begin{array}{ll} \text { Now, } & \angle R O Q+\angle O R Q+\angle O Q R=180^{\circ} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & 2 \angle O Q R=180^{\circ}-150^{\circ} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & 2 \angle O Q R=30^{\circ} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \angle O Q R=15^{\circ} \end{array}\)
Again, \(\begin{array}{ll} \angle O Q R=90^{\circ} \end{array}\)    \([\because O Q \perp Q P]\)
\(\begin{aligned} \Rightarrow \angle O Q R+\angle R Q P=90 \\ \end{aligned}\)^\({\circ}\)
\(\Rightarrow \quad \angle R Q P=90^{\circ}-15^{\circ}=75^{\circ}\)
(iii) (b) We know that angle subtended by an are at centre is double the angle subtended by it at any other part of the circle.
\(\begin{aligned} 2 \angle R S Q & =\angle R O Q \\ \end{aligned}\)
\(\begin{aligned} \angle R S Q & =\frac{1}{2} \times 150^{\circ}=75^{\circ} \end{aligned}\)
(iv) (a) \(\angle\)ORP = 90^\(\circ\) as radius is always perpendicular to the tangent at the point of contact.

(i) (b): \(\angle X A C=45^{\circ}\) 
\(\therefore \quad \angle A C D=45^{\circ}\) [Alternate interior angles]
(ii) (b)
(iii) (c) : \(\text { In } \Delta A C D\)
\(\frac{A D}{D C}=\tan 45^{\circ} \)
\(\Rightarrow \frac{100}{D C}=1 \Rightarrow D C=100 \mathrm{~m}\)
(iv) (d): \(\text { In } \Delta A B D, \frac{A D}{B D}=\tan 30^{\circ}\)
\(\Rightarrow \quad \frac{100}{B D}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow \quad B D=100 \sqrt{3} \mathrm{~m}\)
(v) (a): \(\text { In } \Delta A D C\)
\(\frac{A D}{A C}=\sin 45^{\circ} \Rightarrow \frac{100}{A C}=\frac{1}{\sqrt{2}} \Rightarrow A C=100 \sqrt{2} \mathrm{~m}\)

(i) (a): We have, AB = 9 m, BC = 3\(\sqrt{3}\) m
In  \(\Delta\)ABC, we have
\(\tan A=\frac{B C}{A B}=\frac{3 \sqrt{3}}{9}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow \tan A=\tan 30^{\circ} \Rightarrow \angle A=30^{\circ}\)
(ii) (c): Similarly, \(\tan C=\frac{A B}{B C}=\frac{9}{3 \sqrt{3}}=\sqrt{3}\)
\(\Rightarrow \tan C=\tan 60^{\circ} \Rightarrow \angle C=60^{\circ}\)
(iii) (d): Since \(\sin A=\frac{B C}{A C} \Rightarrow \sin 30^{\circ}=\frac{B C}{A C}\)
\(\Rightarrow \frac{1}{2}=\frac{3 \sqrt{3}}{A C} \Rightarrow A C=6 \sqrt{3} \mathrm{~m}\)
(iv) (b) : \(\because \angle A=30^{\circ}\) [From (1)]
\(\therefore \quad \cos 2 A=\cos \left(2 \times 30^{\circ}\right)=\cos 60^{\circ}=\frac{1}{2}\) 
(v) (b): \(\because \angle C=60^{\circ}\) [Using (2)]
\(\therefore \quad \sin \left(\frac{C}{2}\right)=\sin \left(\frac{60^{\circ}}{2}\right)=\sin 30^{\circ}=\frac{1}{2}\)