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

To find the length of the side BC, we will choose the trigonometric ratio involving BC and the given side AB. Since BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore

\(\begin{aligned} & \frac{\mathrm{AB}}{\mathrm{BC}}=\tan \mathrm{C} \\ \end{aligned}\)
i.e.,   \(\begin{aligned} & \frac{5}{\mathrm{BC}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\ \end{aligned}\)
which gives \(\begin{aligned} & \mathrm{BC}=5 \sqrt{3} \mathrm{~cm} \end{aligned}\)
To find the length of the side AC, we consider
\(\begin{aligned} \sin 30^{\circ} & =\frac{\mathrm{AB}}{\mathrm{AC}} \\ \end{aligned}\)
i.e., \(\begin{aligned} \frac{1}{2} & =\frac{5}{\mathrm{AC}} \\ \end{aligned}\)
i.e., AC = 10 cm
Note that alternatively we could have used Pythagoras theorem to determine the third side in the example above,
i.e., \(\mathrm{AC}=\sqrt{\mathrm{AB}^2+\mathrm{BC}^2}=\sqrt{5^2+(5 \sqrt{3})^2} \mathrm{~cm}=10 \mathrm{~cm} .\)

1268 m

Let aeroplane be at B and let the two ships at C and D, such that their angles of depression from B are \(30°\)and\(60°\), respectively.Then, angles of elevation of D and C from B are \(30°\)and \(60°\)respectively.

We have, AB=1200 m
Here, one side AB is common in both triangles.
Let AC=x m and CD= y m
In \(\Delta BAC\) , we have
\(tan\quad 60°=\frac { A }{ B } \Rightarrow \sqrt { 3 } =\frac { 1200 }{ x } \)
⇒ \(x=\frac { 1200 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 1200\sqrt { 3 } }{ 3 } =400\sqrt { 3 } \)  ....(i)
In \(\Delta BAD\) , We have
\(tan\quad 30°=\frac { AB }{ AD }=\frac { AB }{ DC+CA }\)    [∵ AD=DC+CA]
On putting the value of x from Eq.(i) in Eq. (ii) we get
\(y=1200\sqrt { 3 } -400\sqrt { 3 } \)
⇒ y = 1800\(\sqrt { 3 } \)
⇒ y = 800 x 1.732 \(\left[ \because \sqrt { 3 } =1.732 \right] \)
\(\Rightarrow y=1385.6m\)
Hence, the distance between both ships is 1385. 6m.

We are given two concentric circles C₁ and C₂ with centre O and a chord AB of the larger circle C₁ which touches the smaller circle C₂ at the point P (see Fig). We need to prove that AP = BP.

Let us join OP. Then, AB is a tangent to C₂ at P and OP is its radius. Therefore, by Theorem.
OP \(\perp\) AB
Now AB is a chord of the circle C₁ and OP \(\perp\) AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord,
i.e., AP = BP

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 = \(=\frac{\theta}{360^{\circ}} \times \pi r^2\)
\(\begin{aligned} & =\frac{36^{\circ}}{360^{\circ}} \times \frac{22}{7}\left(\frac{35}{2}\right)^2 \quad\left[\because r=\frac{d}{2}=\frac{35}{2} \mathrm{~mm}\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{10} \times \frac{22}{1} \times \frac{5}{2} \times \frac{35}{2}=\frac{11 \times 35}{2 \times 2}=\frac{385}{4} \mathrm{~mm}^2 \end{aligned}\)

Let AB be the tangent drawn at a point C on the circle with centre O.
To prove Perpendicular at point C passes through the centre O. If possible, let the perpendicular passing through some other point say O'.
Construction Join OC and O' C.

Proof Since, tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(\therefore\) OC \(\perp\) AB \(\Rightarrow\) \(\angle\)OCB = 90^o
Also, \(\angle\)O'CB = 90^o  [as it is supposed that CO' \(\perp\)AB]
\(\therefore\)  \(\angle\)OCB = \(\angle\)O' CB
which is possible only when points O and O' coincide.
So, our assumption is wrong.
Hence, the perpendicular at the point of contact to the tangent to a circle always passes through the centre.
Hence proved.

We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact  Fig. We need to prove that
 
\(\angle \mathrm{PTQ}=2 \angle \mathrm{OPQ}\) 
Let \(\angle \mathrm{PTQ}=\theta\) 
Now, TP = TQ. So, TPQ is an isosceles triangle.
Therefore, \(\angle \mathrm{TPQ}=\angle \mathrm{TQP}=\frac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\frac{1}{2} \theta\) 
Also, \(\angle \mathrm{OPT}=90^{\circ}\) 
So, \(\angle \mathrm{OPQ}=\angle \mathrm{OPT}-\angle \mathrm{TPQ}=90^{\circ}-\left(90^{\circ}-\frac{1}{2} \theta\right)\) 
\(=\frac{1}{2} \theta=\frac{1}{2} \angle \mathrm{PTQ}\) 
This gives \(\angle \mathrm{PTQ}=2 \angle \mathrm{OPQ}\)

LHS = \(\frac { \cos { A } -\sin { A } +1 }{ \cos { A } +\sin { A } -1 } \)
On dividing numerator and denominator by sin A, we get
\(=\frac { \frac { \cos { A } }{ \sin { A } } -\frac { \sin { A } }{ \sin { A } } +\frac { 1 }{ \sin { A } } }{ \frac { \cos { A } }{ \sin { A } } +\frac { \sin { A } }{ \sin { A } } -\frac { 1 }{ \sin { A } } } =\frac { \cot { A } -1+cosecA }{ \cot { A } +1-1cosecA } \) \(\left[ \because \cot { A } =\frac { \cos { \theta } }{ \sin { \theta } } and\frac { 1 }{ \sin { \theta } } =cosec \theta \right] \)
\(=\frac { \cot { A } +cosecA-1 }{ \cot { A } +1-1cosecA } \)
\(=\frac { (\cot { A } +cosecA)-({ cosec }^{ 2 }A-\cot ^{ 2 }{ A } ) }{ \cot { A } +1-1cosecA } \left[ \because 1={ cosec }^{ 2 }A-\cot ^{ 2 }{ A } \right] \)
\(=\frac { (\cot { A } +cosecA)-\left[ (\cot { A } +cosecA)(\cot { A } -cosecA) \right] }{ \cot { A } +1-1cosecA } \) \(\left[ \because \quad { a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b) \right] \)
\(=\frac { (\cot { A } +cosecA)-\left[ 1-(cosecA-\cot { A } ) \right] }{ \cot { A } +1-1cosecA } \)
\(\quad =\frac { (\cot { A } +cosecA)-\left[ 1-cosecA+\cot { A } \right] }{ \cot { A } +1-1cosecA } \)
\(=cosecA+\cot { A } =RHS\)
Hence proved.

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 AB = 30 m be the height of the building and DC = 1.5 m be the height of the boy. The point D be the boy's eye.
Draw the line DF || CA.
Then, CD = AF = 1.5 m
The angle of elevation is \(\angle\)BDF = 30°.
Let he walked DE = xm towards the building. Then, the angle of elevation is \(\angle\)BEF = 60°.
Let EF = y m
Now, BF = AB - AF = 30 - 1.5 = 28.5 m.

In right angled \(\Delta\)BFD,
\(\begin{aligned} & \tan 30^{\circ}=\frac{P}{B}=\frac{B F}{D F}=\frac{B F}{D E+E F} \quad[\because D F=D E+E F] \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{28.5}{x+y} \quad\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right] \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad x+y=28.5 \sqrt{3} \mathrm{~m} \\ \end{aligned}\)    ...(i)
and in right angled \(\Delta\)BFE,
\(\begin{aligned} \tan 60^{\circ} & =\frac{B F}{E F} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \sqrt{3} & =\frac{28.5}{y} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right] \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad y & =\frac{28.5}{\sqrt{3}} \mathrm{~m} \end{aligned}\)
On putting \(y=\frac{28.5}{\sqrt{3}} \text { in Eq. (i), we get } x+\frac{28.5}{\sqrt{3}}=28.5 \sqrt{3}\)
\(\begin{aligned} \Rightarrow \quad x & =28.5\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=28.5\left(\frac{3-1}{\sqrt{3}}\right) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{28.5 \times 2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \end{aligned}\)     [rationalising]
\(\begin{aligned} =\frac{57 \sqrt{3}}{3}=19 \sqrt{3} \mathrm{~m} \end{aligned}\)
Hence, the distance he walked towards the building, is 19\(\sqrt3\)m.

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.

\(L H S=\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}\)
\(=\frac{\cos A\left(\frac{1}{\sin A}-1\right)}{\cos A\left(\frac{1}{\sin A}+1\right)}=\frac{\left(\frac{1}{\sin A}-1\right)}{\left(\frac{1}{\sin A}+1\right)}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\operatorname{RHS}\)

Let AB be the tower of height x m, and CD be the tower of height y m.

Let E be the mid-point of the line AC.
Then ∠AEB = 30° and ∠CED = 60°.
Also, AE = EC = a m (let)
In fight angled ΔBAE tan 30° \(=\frac{P}{B}=\frac{A B}{A E}=\frac{x}{a}\)
and in right angled ΔDCE, tan 60° \(=\frac{D C}{C E}=\frac{y}{a}\) 
1 : 3

From every point of a circle, we can draw a tengent. Therefore, infinite tangents can be drawn.

In art. \(\triangle\)ABC, \(\angle\)B = 90°
\(\therefore\)   AB = 2 m, BC = \(2\sqrt{3}\)m
Let  \(\angle\)ACB = \(\theta\)
\(\therefore\)  tan \(\theta={{AB}\over{BC}}={{2}\over{2\sqrt{3}}}\)
tan \(\theta ={{1}\over{\sqrt{3}}}\Rightarrow{\theta}=30°\)

Given, sector angle, \(\theta\)= 80°
and distance or radius, r = 16.5 km
\(\begin{aligned} \therefore \text { Area of sector } & =\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{80^{\circ}}{360^{\circ}} \times 3.14 \times(16.5)^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{2 \times 3.14 \times 272.25}{9} \end{aligned}\)
\(\begin{aligned} =\frac{1709.73}{9} \end{aligned}\)
= 189.97 km²
which is the required area of the sea over which the ships are warned.

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
\(=\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(45)^2[\because r=45, \text { given }]\)
\(\begin{aligned} & =\frac{22}{7 \times 8}(45)^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22275}{28} \mathrm{~cm}^2 \end{aligned}\)

(i) (a): Area of sector ODCO \(=\frac{1}{4} \pi r^{2}\)
\(=\frac{1}{4} \times \frac{22}{7} \times 14 \times 14=154 \mathrm{~cm}^{2}\)
(ii) (b): Area of \(\triangle A O B=\frac{1}{2} \times O A \times O B=\frac{1}{2}(20 \times 20)\)
= 200 cm²
(iii) (a) : Area of region which is golden plated
= area of \(\Delta\)OAB - area of sector ODCO.
= 200 - 154 = 46 cm²
\(\therefore\) Total cost of golden plating = Rs (6 x 46) = Rs 276
(iv) (c): Area of major sector = area of circle - area of minor sector
\(=\pi r^{2}-\frac{1}{4} \pi r^{2}=\frac{3 \pi r^{2}}{4}=\frac{3}{4} \times \frac{22}{7} \times 14 \times 14=462 \mathrm{~cm}^{2}\)
(v) (d): Length of arc DC \(=\frac{90^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 14\)
= 22cm

(i) (a)
(ii) (d)
(iii) (b)
(iv) (a)
(v) (c)

(i) (c): \(\text { In } \Delta O P Q\), we have
\(\tan 60^{\circ}=\frac{P Q}{P O} \)
\(\Rightarrow \sqrt{3}=\frac{20}{P O} \)
\(\Rightarrow P O=\frac{20}{\sqrt{3}} \mathrm{~m}\)
(ii) (b): In \(\Delta\)ORS, we have
\(\tan 30^{\circ}=\frac{R S}{O R} \Rightarrow \frac{1}{\sqrt{3}}=\frac{20}{O R} \Rightarrow O R=20 \sqrt{3} \mathrm{~m}\)
(iii) (d): Clearly, width of the road = PR
\(\begin{array}{l} =P O+O R=\left(\frac{20}{\sqrt{3}}+20 \sqrt{3}\right) \mathrm{m} \\ =20\left(\frac{4}{\sqrt{3}}\right) \mathrm{m}=\frac{80}{\sqrt{3}} \mathrm{~m}=46.24 \mathrm{~m} \end{array}\)
(iv) (a): \(\text { In } \Delta O P Q \text { , if } \angle P O Q=45^{\circ} \text { , then }\)
\(\tan 45^{\circ}=\frac{P Q}{P O} \Rightarrow 1=\frac{20}{P O} \Rightarrow P O=20 \mathrm{~m}\)
(v) (b)

(i) (d) : \(\text { In } \Delta A P Q, \tan \theta=\frac{A Q}{P Q}=\frac{1.2}{1.6}=\frac{3}{4}\)
(ii) (d) : \(\text { In } \Delta P B Q, \cot B=\frac{Q B}{P Q}=\frac{3}{1.6}=\frac{15}{8}\) ...(i)
(iii) (c): \(\text { In } \Delta A P Q, \tan A=\frac{P Q}{A Q}=\frac{1.6}{1.2}=\frac{4}{3}\) ...(ii)
(iv) (d): We have, tan²A + 1 = sec²A
\(\Rightarrow \sec A =\sqrt{\left(\frac{4}{3}\right)^{2}+1} \)
\(=\sqrt{\frac{16}{9}+1}=\sqrt{\frac{25}{9}}=\frac{5}{3}\)
(v) (a): Since \(\operatorname{cosec} B=\sqrt{\cot ^{2} B+1}\)
\(\begin{array}{l} =\sqrt{\left(\frac{15}{8}\right)^{2}+1} \\ =\frac{17}{8} \end{array}\)