\(L.H.S=\left( \sin { \theta } +cosec\theta \right) ^{ 2 }+\left( \cos { \theta } +\sec { \theta } \right) ^{ 2 }\)
\(=\sin ^{ 2 }{ \theta } +{ cosec }^{ 2 }\theta +2\sin { \theta } cosec\theta +\cos ^{ 2 }{ \theta } +\sec ^{ 2 }{ \theta } +2\cos { \theta } \sec { \theta } \)
\(=\left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) +{ cosec }^{ 2 }\theta +2\sin { \theta } \times \frac { 1 }{ \sin { \theta } } +\sec ^{ 2 }{ \theta } +2\cos { \theta } \times \frac { 1 }{ \cos { \theta } } \)
\(=1+\left( 1+\cot ^{ 2 }{ \theta } \right) +2+\left( 1+\tan ^{ 2 }{ \theta } \right) +2\)
\(=7+\tan ^{ 2 }{ \theta } +\cot ^{ 2 }{ \theta } \)
= R.H.S

(i) Let the assumed mean, A = 1400 and h = 400
       Calculation of Mean

We have A = 1400, h = 400,  \(\sum { f_{ i } }{ u_{ i } }\) = -1395 and N = 1343
Mean = A + \(h\left\{ \frac { 1 }{ N } \sum { f_{ i } } { u }_{ i } \right\} \)
 = 1400 + 400\(\times\) \(\left( \frac { -1395 }{ 1343 } \right) \)
= 1400 - 415.49 = 984.51
(ii) Mean by assumed mean method.
(iii) Villages are necessary to keep a balance in nature.

It can be observed from the given data that the maximum class frequency is 40, belonging to 1500 − 2000 intervals.
Therefore, modal class = 1500 − 2000
Lower limit (l) of modal class = 1500
Frequency (f₁) of modal class = 40
Frequency (f₀) of class preceding modal class = 24
Frequency (f₂) of class succeeding modal class = 33
Class size (h) = 500
\(\text { Mode }=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \)
\(=1500+\left(\frac{40-24}{2(40)-24-33}\right) \times 500 \)
\(=1500+\left(\frac{16}{80-57}\right) \times 500\)
=1500+8000/23
Therefore, modal monthly expenditure was Rs 1847.83.
To find the class mark, the following relation is used.
\(Class mark =\frac{\text { Upper class limit + Lower class limit }}{2}\)
Class size (h) of the given data = 500
Taking 2750 as assumed mean (a), d_i, u_i, and f_iu_iare calculated as follows
Taking 2750 as assumed mean (a), d_i, u_i, and f_iu_iare calculated as follows.

From the table, we obtain
\(\sum f_{i}=200 \)
\(\sum f_{i} u_{i}=-35 \)
\(\text { Mean } \bar{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) x h \)
\(\bar{x}=2750+\left(\frac{-354}{200}\right) \times 500\)
= 2750 - 87.5
= 2662.5
Therefore, mean monthly expenditure was Rs 2662.50.

1732m

Let BC = x m be the width of the canal and AB = h m be the height of the tower.
Given, \(\angle\)ACB = 60\(\unicode{xb0} \) and  \(\angle\)ADB = 30\(\unicode{xb0} \)
In right angled \(\Delta\)ABD, 
\(\tan 30^{\circ}=\frac{P}{B}=\frac{A B}{D B}\)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{D C+C B}\)

\(\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}} \text { and } D B=D C+C B\right]\)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{20+x} \Rightarrow 20+x=\sqrt{3} h\)  ...(i)
and in right angled \(\Delta\)ABC,
\(\tan 60^{\circ}=\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{h}{x} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right]\)
\(\Rightarrow \quad h=\sqrt{3} x\)                                      ..(ii)
On putting h = \(\sqrt{3}\)x in Eq. (i), we get
20 + x = \(\sqrt{3}\) (\(\sqrt{3}\)x)
\(\Rightarrow\) 20 + x = 3x
\(\Rightarrow\) 2x = 20 \(\Rightarrow\) x = 10 m
On putting x = 10 m in Eq. (ii), we get h = \(\sqrt{3}\)(10)
\(\Rightarrow\) h = 10\(\sqrt{3}\) m
Hence, the height of the tower is 10\(\sqrt{3}\) m and width of the canal is 10 m.

Let PQ and PR be two tangents drawn from an external point P to a circle with centre O.

To prove \(\angle\)QOR = 180° – \(\angle\)QPR
or \(\angle\)QOR + \(\angle\)QPR = 180°
Proof In \(\Delta\)OQP and \(\Delta\)ORP,
PQ = PR   [\(\because\) tangents drawn from an external point are equal in length]
OQ = OR      [radii of circle]
OP  = OP   [common sides]
\(\therefore\)  \(\Delta\)OQP \(\cong\) \(\Delta\)ORP   [by SSS congruence rule]
Then, \(\angle\)QPO = \(\angle\)RPO    [by CPCT]
and \(\angle\)POQ = \(\angle\)POR    [by CPCT]
\(\left.\begin{array}{ll} \Rightarrow & \angle Q P R=2 \angle O P Q \\ \text { and } & \angle Q O R=2 \angle P O Q \end{array}\right\}\)    ...(i)
Now, in right angled \(\Delta\)OQP, \(\angle\)QPO + \(\angle\)QOP = 90°
\(\Rightarrow\)  \(\angle\)QOP = 90° - \(\angle\)QPO
\(\Rightarrow\)  2\(\angle\)QOP = 180°- 2 \(\angle\)QPO
[multiplying both sides by 2]
\(\Rightarrow\) \(\angle\)QOR = 180° - \(\angle\)QPR      [from Eq. (i)]
\(\Rightarrow\)   \(\angle\)QOR + \(\angle\)QPR = 180°      Hence proved.

Let original total expenditure of mess was Rs x
\(\Rightarrow\) Average expenditure per student = Rs \({ { x}\over{ 35} }\)
New expenditure of mess = Rs x + 42
Average expenditure per student = \({ { x+42 }\over{42 } }={ {x }\over{42 } }+1\)
\(\Rightarrow\)  \({ { x}\over{ 42} }+1={ { x}\over{ 35} }-1\Rightarrow1+1={ { x}\over{35 } }-{ {x }\over{ 42} }\)
\(\Rightarrow\) 2 = \({ {6x-5x }\over{ 210} }\Rightarrow x=420\)
Original expenditure of mess was Rs 420

From the triangle ABC
\({h\over 28.2}=tan60^0\)
h = 28.2 x tan 60°
= 28.2 x \(\sqrt3\)
= 28.2\(\sqrt3\) + 1.2m
Height ofthe tower = AC+CD
= 50 m

Let AB = 20 m be the height of tower. Let the ball lying on the ground at point C.
∴ ㄥACB = 60\(\unicode{xb0} \)
Consider rt. angled ΔABC, we have
\(\frac { AB }{ AC } \) = tan 60\(\unicode{xb0} \)
\(\frac { 20 }{ BC } =\sqrt { 3 } \)
⇒ BC = \(\frac { 20 }{ \sqrt { 3 } } =\frac { 20\sqrt { 3 } }{ 3 } \)
=\(\frac { 20\times 1.73 }{ 3 } \)
=11.53 m.
∴ The distance between the foot of the tower and the ball is 11.53 m.

Area of sector OCDE \(=\frac{60^{\circ}}{360^{\circ}} \pi r^{2}\)
\(\begin{array}{l} =\frac{1}{6} \times \frac{22}{7} \times 6 \times 6 \\ =\frac{132}{7} \mathrm{~cm}^{2} \end{array}\)
Area of triangle OAB \(=\frac{\sqrt{3}}{4}(12)^{2}=\frac{\sqrt{3} \times 12 \times 12}{4}=36 \sqrt{3} \mathrm{~cm}^{2}\) 
Area of circle \(=\pi r^{2}=\frac{22}{7} \times 6 \times 6=\frac{792}{7} \mathrm{~cm}^{2}\)
Area of shaded region = Area of ΔOAB + Area of circle − Area of sector OCDE
\(\begin{array}{l} =36 \sqrt{3}+\frac{792}{7}-\frac{132}{7} \\ =\left(36 \sqrt{3}+\frac{660}{7}\right) c m^{2} \end{array}\)

In given figure,
sin 45^o = \(\frac { BC }{ AC } \Rightarrow \frac { 1 }{ \sqrt { 2 } } =\frac { BC }{ 150 } \)
BC = \(\frac { 150 }{ \sqrt { 2 } } =75\sqrt { 2 } \)m

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

Join OL
OL RS
Also    OL  = OM   [Radii of the same circle]
ஃ ⇒ OLM = 30^o
⇒ o - 30^o = 60^o

Speed of ship \(=36 \mathrm{~km} / \mathrm{h}\)
Time taken \(=2 \mathrm{~min}\)
Therefore Distance covered by ship in \(2 \mathrm{~min}=36 \times \frac{2}{60}=1.2 \mathrm{~km}\)
Distance of ship from foot of lighthouse =1.3-1.2
\(=0.1 \mathrm{~km} =100 \mathrm{~m}\)
Let angle of elevation of the ship to the top of the lighthouse \(=y^{\circ}\)
Then, \(\tan y=\frac{173.2}{100}=1.732\)
\(\Rightarrow \quad \tan y=\sqrt{3} {[\because 1.732=\sqrt{3}]} \)
\( \Rightarrow \quad y=60^{\circ} \)
\({\left[\because \tan \sqrt{3}=\tan 60^{\circ}\right]}\)
Now, \(y=x^{\circ} \quad[\because\) alternate interior angles are equal ]
Thus, \(x^{\circ}=60^{\circ}\)

\(\frac { \sec ^{ 2 }{ \left( { 90 }^{ ° }-\theta \right) -\cot ^{ 2 }{ \theta } } }{ 2\left( \sin ^{ 2 }{ { 25 }^{ ° } } +\sin ^{ 2 }{ { 65 }^{ ° } } \right) } -\frac { 2\cos ^{ 2 }{ { 60 }^{ ° }\tan ^{ 2 }{ { 28 }^{ ° }\tan ^{ 2 }{ { 62 }^{ ° } } } } }{ 3\left( \sec ^{ 2 }{ { 43 }^{ ° } } -\cot ^{ 2 }{ { 47 }^{ ° } } \right) } \)
\(=\frac { \left( { cosec }^{ 2 }\theta -\cot ^{ 2 }{ \theta } \right) }{ 2\left( \sin ^{ 2 }{ { 25 }^{ ° } } +\cos ^{ 2 }{ { 25 }^{ ° } } \right) } -\frac { 2\times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \tan ^{ 2 }{ { 28 }^{ ° } } \times \cot ^{ 2 }{ { 28 }^{ ° } } }{ 3\left[ \sec ^{ 2 }{ { 43 }^{ ° } } -\tan ^{ 2 }{ { 43 }^{ ° } } \right] } \)
\(=\frac { 1 }{ 2\times \left( 1 \right) } -\frac { \frac { 1 }{ 2 } \times \tan ^{ 2 }{ { 28 }^{ ° }\times \frac { 1 }{ \tan ^{ 2 }{ { 28 }^{ ° } } } } }{ 3 } \)
\(=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\)

Let the height be 'h' m and breadth be 'b' m.
In \(\Delta ABC\),
\(\frac { h }{ b } =tan60°=\sqrt { 3 } \)
h=\(\sqrt { 3 } b\)
In \(\Delta ABD\), \(\frac { h }{ b+20 } =tan30°=\frac { 1 }{ \sqrt { 3 } } \)
h=\(\frac { b+20 }{ \sqrt { 3 } } \)
b\(\sqrt { 3 } \)=\(\frac { b+20 }{ \sqrt { 3 } } \)
3b = b + 20 => 2b = 20 => b = 10m
h=b\(\sqrt { 3 } \)=10 x 1.73=17.3 m
Height of tree is 17.3m and breadth of river is 10m.

N=500, \(\frac { N }{ 2 } =250\)

Median class = 1000-2000
Median = l + \(\frac { \frac { N }{ 2 } -c.f }{ f } \times h\)

\(=1000+\frac { 250-150 }{ 200 } \times 1000\)
=1000+500=1500
Median Expenditure = Rs 1500 /week

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.

Join OR and OS and let r be the radius of circle

BC = 38 cm, CD = 28 cm 
BP 25 cm = BQ    [tangent from external point] 
CQ = BC- BQ = 38 - 25= 13 cm 
∴ CR = CQ 13 cm
RD = CD - CR = 28 -13 = 15 cm 
Quad. DROS is a square (3 angles are 90°) 
∴ r = DR = 15 cm 

Here, AC = 24 cm, BC = 10 cm and AOB is a diameter. 
Since angle in semicircle is a right angle i.e., \(\angle \)ACB = 90°
ஃ  AB = \(\sqrt{AC^2+BC^2}\)
= \(\sqrt{24^2+10^2}\) = \(\sqrt{576+100}\)
= \(\sqrt{676}\) = 26 cm
ஃ RAdius, area of circle = \(26\over2\) = 13 cm
Now, Area of semicircle
= Area of semicircle - Area of \(\triangle\)ABC
 = \(1\over2\)\(\pi\)r² - \(1\over2\) x BC x BC
= \(1\over2\) x 3.14 x 13 x 13 - \(1\over2\) x 24 x 10
= \(530.66\over2\) - 120
= 265.33 - 120 = 145.33 cm²

If Assertion is incorrect but Reason is correct.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(i) (c) From table, we see that maximum frequency is 23.Therefore, modal class is 35-45.
Here, l=35, f₁ = 23, f₀ = 21, f₂ = 14 and h = 10
\(\begin{aligned} \therefore \text { Mode } & =l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h \\ \end{aligned}\)
\(\begin{aligned} & =35+\frac{23-21}{46-21-14} \times 10 \\ \end{aligned}\)
\(\begin{aligned} & =35+\frac{20}{11}=35+1.82=36.82 \end{aligned}\)
(ii) (d) Since, class interval 35-45 has highest frequency i.e. 23, So it is modal class.
\(\therefore\) Upper limit of modal class is 45.
(iii) (d) Calculation of mean

\(\therefore\) \(\text { Mean, } \bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{2830}{80}=35.375=35.4\)
(iv) (a) Here, maximum frequency is 42 and the class corresponding to this frequency is 35 – 45. So, the modal class is 35-45.
\(\therefore\) l = 35, f₁ = 42, f₀ = 10, f₂ = 24 and h = 10
Now, mode = \(\begin{aligned} & =l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h \\ \end{aligned}\)
\(\begin{aligned} & =35+\frac{42-10}{2 \times 42-10-24} \times 10 \\ \end{aligned}\)
\(\begin{aligned} & =35+\frac{32}{50} \times 10=35+6.4=41.4 \end{aligned}\)
(v) (b) The cumulative frequency table for given data is

Here, N = 112
\(\therefore \quad \frac{N}{2}=\frac{112}{2}=56\)

Since, the cumulative frequency just greater that 56 is 76 and the corresponding class interval is 35-45.
\(\therefore\) Median class = 35 - 45,
l = 35, cf = 34, h = 10 and f = 42
\(\begin{aligned} \therefore \text { Median } & =l+\frac{\left(\frac{N}{2}-f\right)}{f} \times h=35+\frac{56-34}{42} \times 10 \\ \end{aligned}\)
\(\begin{aligned} & =35+\frac{22}{42} \times 10=35+5.24=40.24 \end{aligned}\)

(a) (iv) NS and NR are both tangent to the circle
So, \(N S \perp S A\) and \(N R \perp R A\) 
[ \(\therefore\)  Tangent to a circle is perpendicular to the radius through the point of contact ]
\(\therefore \angle N S A=90^{\circ}\) 
(b) (ii) 
\(\angle N S A=90^{\circ} \quad \Rightarrow N A^{2}=N S^{2}+S A^{2}\) [ By Pythagora's Theorem]
\(\Rightarrow N S=\sqrt{N A^{2}-S A^{2}}=\sqrt{13^{2}-5^{2}}=\sqrt{169-25}\) 
\(=\sqrt{144}=12 \mathrm{~m}\) 
(c) (iiii)
NR = NS = 12m
Tangents drawn from an external point are equal
\(\therefore N R=12 \mathrm{~m}\) 
(d) (iv)
\(\angle S N R+\angle R A S=180^{\circ} \Rightarrow \angle R A S=180^{\circ}-\angle S N R\) 
[\(\because\) The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact to the centre]
\(\Rightarrow \angle R A S=180^{\circ}-\theta\) 
(e) (i) 
\(\angle R A S=180^{\circ}-\theta\) [ Calulated bove]
Now, \(\angle N A S=\angle N A R=\frac{1}{2} \angle R A S\) 
[\(\because\) If two tangents are drawn from an external point, then they extend equal angles at the centre]
\(\Rightarrow \angle N A S=\frac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\frac{\theta}{2}\)

(i) (c)
(ii) (a): Let AB be the height of pedestal.
\(\text { In } \Delta A B C, \)
\(\tan 30^{\circ}=\frac{A B}{B C} \)
\(\Rightarrow \quad A B=\frac{25}{\sqrt{3}}=\frac{25}{1.73}=14.45 \mathrm{~m}\)

(iii) (d): Let x be the distance between Arun and Pankaj.

\(\text { In } \Delta A B \dot{D}, \tan 45^{\circ}=\frac{A B}{B D}\)
\(\Rightarrow B D=30 \mathrm{~m}\)
\(\text { Now, in } \Delta \overline{A B C} \text { , }\)
\(\tan 30^{\circ}=\frac{A B}{B C}\)
\(\Rightarrow \frac{30}{30+x}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow x=30(\sqrt{3}-1)=30 \times 0.73=21.9 \mathrm{~m}\)
(iv) (c): \(\text { In } \triangle A R S\)

\(\sin 30^{\circ}=\frac{R S}{A S} \)
\(\Rightarrow \quad \frac{12}{A S} =\frac{1}{2} \Rightarrow A S=12 \times 2=24 \mathrm{~m}\)
(v) (c): \(\text { In } \Delta A B D, \frac{A B}{B D}=\tan 60^{\circ}\)

\(\begin{array}{l} \Rightarrow \frac{A B}{15}=\sqrt{3} \\ \Rightarrow A B=15 \times 1.73=25.95 \mathrm{~m} \end{array}\)