Given, \(\sqrt { 2 } \sin { \theta } =1\)
\(\sin { \theta } =\frac { 1 }{ \sqrt { 2 } } =\sin { { 45 }^{ ° } } \)
\(\therefore \quad \theta ={ 45 }^{ ° }\)
Now \(\sec ^{ 2 }{ \theta } -cosec^{ 2 }\theta \) = \(\sec ^{ 2 }{ { 45 }^{ ° } } -cosec^{ 2 }{ 45 }^{ ° }\)
\(={ \left( \sqrt { 2 } \right) }^{ 2 }-{ \left( \sqrt { 2 } \right) }^{ 2 }\)
= 2 - 2
= 0 

\(\tan { { 30 }^{ ° } } =\frac { OA }{ AP } \)
\(\Rightarrow \quad \frac { 1 }{ \sqrt { 3 } } =\frac { \sqrt { 3 } }{ AP } \)
\(\Rightarrow \quad AP=\sqrt { 3 } \times \sqrt { 3 } =\) 3 cm

Points are (-4, -7) and (0, -7)
Distance = \(=\sqrt { { \left( 0+4 \right) }^{ 2 }+{ \left( -7+7 \right) }^{ 2 } } =\sqrt { { 4 }^{ 2 }+0 } =\sqrt { 16 } \) = 4 units

Here, a = 6, d = 6, n = 5
\(\because\) S = n/2[2a+(n-l)d]
S₅ = 5/2[2x6+(5-1)(6)]
= 5/2[12+4x6]
= 5/2[12+24] = 5/2[36]
= 5x18 = 90

Since, - 1 is a zero of the polynomial
p(x) = kx² - 4x + k,
then p(-1) = 0
\(\therefore\)  k(-1)² - 4 (- 1) + k = 0
\(\Rightarrow\)  k + 4 + k = 0
\(\Rightarrow\)  2k + 4 = 0
\(\Rightarrow\)  2k = - 4
Hence, K = - 2

32

1. Draw a circle of centre O and radius 4 cm.
2. Take a point P outside the circle and draw a secant PAB, intersect~ng the circle at A and B.
3. Produce AP to C such that AP = CP.
4. Draw a semi-circle with CB as diameter.
5. Draw \(PD\bot CB\)  intersecting the semi-circle at D.
6. With P as centre and PD as radius, draw arcs to intersect the given circle at T and T'.
7. Join PT and PT'. Thus, PT and PT' are the required tangents.

The common point of the tangent to a circle and the circle is called the point of contact.

Number of bad eggs\(=400\times0.035=400\times{35\over1000}\)
= 14

Here, \(\frac { -2+8 }{ 2 } =\frac { 6+x }{ 2 } \Rightarrow x=0\)
and    \(\frac { 3+3 }{ 2 } =\frac { 7+y }{ 2 } \Rightarrow y=-1\)
Thus, the fourth vertex D of parallelogram ABCD is D(0,-1).

Let AB be tower of height h m and C is a point on the ground such that BC=60 m

As ㄥACB=600 and AB=h m
Consider rt. angled ΔABC, we have
tan 600=\(\frac { AB }{ BC } \)
⇒ \(\sqrt { 3 } =\frac { h }{ 60 } \)
⇒ h=60\(\sqrt { 3 } \) m.
Hence, the height of the tower is 60 \(\sqrt { 3 } \) m.

The shape of the well will be cylindrical.
Depth (h) of well = 20 m
Radius (r) of circular end of well = 7/2 m
Area of platform = Length × Breadth = 22 × 14 m²
Let height of the platform = H
Volume of soil dug from the well will be equal to the volume of soil scattered on the platform.
Volume of soil from well = Volume of soil used to make such platform
Π x r² x h = Area of platform x Height of Platform
\(\begin{array}{l} \Pi \times\left(\frac{7}{2}\right)^{2} \times 20=22 \times 14 \times H \\ \therefore H=\frac{22}{7} \times \frac{49}{4} \times \frac{20}{22 \times 14}=\frac{5}{2} m=2.5 \mathrm{~m} \end{array}\)
Therefore, the height of such platform will be 2.5 m.

We have :
a₂ – a₁ = 11 – 5 = 6,
a₃ – a₂ = 17 – 11 = 6,
a₄ – a₃ = 23 – 17 = 6
As a_{k + 1} – a_k is the same for k = 1, 2, 3, etc., the given list of numbers is an AP.
Now, a = 5 and d = 6.
Let 301 be a term, say, the nth term of this AP.
We know that
a_n = a + (n – 1) d
So,   301 = 5 + (n – 1) x 6
i.e.,   301 = 6n – 1
So,  \(n=\frac{302}{6}=\frac{151}{3}\)
But n should be a positive integer . So, 301 is not a term of the given list of numbers.

∴ AB'C' is the required triangle.

If terms are in AP, then 13 - (2p + 1) = (5p - 3) - 13
\(\Rightarrow\)   13 - 2p -1 = 5p - 3 - 13
\(\Rightarrow\) 28 = 7p
\(\Rightarrow\) p = 4

AB² + AC² = 2AD² + 2(\(\frac {1}{2}\)BC)²

Draw AE \(\bot\) BC
In \(\triangle\)ABE,
AB² = AE² + BE² (Pythagoras theorem)
AB² = AD² - DE² + (BD - DE)²   
= AD²- DE² + BD² + DE²-2BD x DE
AB² = AD² + BD² - 2BD x DE ....(i)
In \(\triangle\)AEC,
AC² = AE² + EC²
\(\Rightarrow\) AC² = (AD² - ED²) + (ED + DC)²
= AD²-ED² + ED²+ DC² +2ED x DC
\(\Rightarrow\) AC² = AD² + CD² + 2ED x CD
\(\Rightarrow\) AC2 = A02 + OC2 + 20C x DE ....(ii) 
Adding eqns. (i) & (ii),
AB² + AC² = 2(AD² + BD²)  (\(\therefore\)BD = DC)
\(=2\left[ { AD }^{ 2 }+\left( \frac { 1 }{ 2 } { BC }^{ 2 } \right) \right] \)(as BD = 1/2BC)
or AB² + AC² = 2(AD² + BD²) 
Hence proved .

\(50\sqrt { 3 } cm^{ 2 }\) 

Let a be the first term and d be the common difference of given AP.
According to the question,
\(S_{ 1 }=S_{ n }=\frac { n }{ 2 } \left[ 2a+(n-1)d \right] \\ S_{ 2 }=S_{ 2n }=\frac { 2n }{ 2 } \left[ 2a+(2n-1)d \right] \)
and \(S_{ 3 }=S_{ 3n }=\frac { 3n }{ 2 } \left[ 2a+(3n-1)d \right] \)
Now, \(S_{ 2 }-S_{ 1 }=\frac { 2n }{ 2 } \left[ 2a+(n-1)d \right] -\frac { n }{ 2 } \left[ 2a+(n-1)d \right] \)
\(=\frac { n }{ 2 } \left[ 2\left\{ 2a+(2n-1d \right\} -\left\{ 2a+(n-1d \right\} \right] \\ =\frac { n }{ 2 } \left[ 2a+(3n-1)d \right] \\ \therefore \ 3\left( S_{ 2 }-S_{ 1 } \right) =\frac { 3n }{ 2 } \left[ 2a+(3n-1)d \right] \\ \Rightarrow \ 3\left( S_{ 2 }-S_{ 1 } \right) =S_{ 3 }\ or S_{ 3 }=3\left( S_{ 2 }-S_{ 1 } \right) \)

Let r and R be inner and outer radii respectively,
then, \(2 \pi r=88\)
and \(\pi\left[R^{2}-r^{2}\right]=346.5\)
= \(17.5 \ cm\)
 

(i)Total number of cards in the box
=100+200+50=350
Number of blue cards=50
P(a blue card)\(={50\over 350}={1\over7}\)
(ii)Number of yellow cards=200
Number of non yellow cards =350-200=150
P(not a yellow card)\(={150\over350}={3\over 7}\)
(iii)Number of yellow or blue cards in the box = 200+50=250
Number of neither yellow nor blue cards = 350-250=10
P(neither a yellow nor a blue card)
\(={100\over 350}={2\over 7}\)

∵ A(0,2) is equidistant from the points B(3,p) and C(p,5).
∴ AB = AC ⇒ AB² = AC²
⇒ (0-3)²+(2-p)²=(0-p)²+(2-5)²
⇒ 9+4-4p+p² = p²+9
⇒ 4p=4 ⇒ p=1
Point B is (3,1)
∴ AB = \(\sqrt { { (0-3) }^{ 2 }+{ (2-1) }^{ 2 } } \)
=\(\\ =\sqrt { 9+1 } =\sqrt { 10 } \) units

Let CD be a tower of height x m and BC is a pole and ㄥBAD=600 and ㄥPCA=450 ㄥPCA=ㄥCAD=45⁰
In right ΔCDA, \(\frac { CD }{ DA } \)=tan 45^o
⇒ \(\frac { x }{ DA } \)=1 ⇒ DA= x m
In right ΔBDA, \(\frac { BD }{ DA } \)=tan 60^o
⇒ \(\frac { 5+x }{ x } =\sqrt { 3 } \)
⇒ 5+x=\(\sqrt { 3 } \)x ⇒ 5=(\(\sqrt { 3 } \)-1)x
⇒ x=\(\frac { 5 }{ \sqrt { 3 } -1 } m=\frac { 5(\sqrt { 3 } +1) }{ 2 } m=\frac { 5(1.73+1) }{ 2 } m=\frac { 13.65 }{ 2 } \)m=6.82 m

Table for the given data is

We have, a = 0.10, h = 0.04,
N = 30 and \(\Sigma\)f_iu_i = -1
By step deviation method,
Mean \((\bar{x})=a+\left(\frac{\Sigma f_i u_i}{N}\right) \times b=0.10+\left(\frac{-1}{30}\right) \times 0.04\)
\(=0.10-\frac{0.04}{30}=0.10-0.001=0.099\)
Hence, the mean concentration of SO₂ in air is 0.099 ppm.

LHS = \(\cot ^{ 2 }{ A } \left[ \frac { \sec { A } -1 }{ 1+\sin { A } } \right] +\sec ^{ 2 }{ A } \left[ \frac { \sin { A } -1 }{ 1+\sec { A } } \right] \)
\(=\frac { \cot ^{ 2 }{ A } (\sec { A } -1)(1+\sec { A } )+\sec ^{ 2 }{ A } (\sin { A } -1)(1+\sin { A } ) }{ (1+\sin { A } )(1+\sec { A } ) } \)
\(=\frac { \cot ^{ 2 }{ A } (\sec ^{ 2 }{ A } -1)+\sec ^{ 2 }{ A } (1-\sin ^{ 2 }{ A } ) }{ (1+\sin { A } )(1+\sec { A } ) } \)
\(=\frac { \cot ^{ 2 }{ A } .\tan ^{ 2 }{ A } -\sec ^{ 2 }{ A } \cos ^{ 2 }{ A } }{ (1+\sin { A } )(1+\sec { A } ) } \)
\(=\frac { 1-1 }{ (1+\sin { A } )(1+\sec { A } ) } =0\)
= RHS
Hence proved. 

Total number of stacks = 9. Characteristics of library are
(i) It is an organised collection of information and resources that are useful to gain knowledge.
(ii) It provides a place, where one can get a data related to all fields as academics, corporations, law medical, etc.

Remaining cards of spade=13-3=10
So, favourable outcomes=10,
i.e. n(E₄)=10
∴ P(getting a spade)​​​​​​​\(=\frac { 10 }{ 49 } \)

Given equation is 
\(3x^{ 2 }+5x+2=0\)
Let  \(p(x)=3x^{ 2 }+5x+2\)
On puttings x=-3 in p(x) ,we get
\(p(-3)=3(-3)^{ 2 }+5(-3)+2\)
 =27-15+2=\(14\neq 0\)
Hence,-3 is not a solution of the given equation.

Rs. 7040.00

Let h m be the height of the tower AB. S and T are two distinct points from its foot, such that AS = s and AT =t 

Consider rt. \(\angle ed\Delta SAB,\) we obtain
\(\frac { AB }{ AS } =tan\theta \quad \) 
\(\frac { h }{ s } =tan\theta \quad \) 
\(h=stan\theta \) 
Again, consider rt.\(\angle ed\Delta TAB,\) we obtain 
\(\frac { AB }{ AT }=tan\left( { 90 }^{ o }-\theta \right) \) 
\(\frac { h }{ t } =cot\theta \) 
\(h=t\times \frac { 1 }{ tan\theta } \) 
Multiply (i) and (ii) we have 
\({ h }^{ 2 }=stan\theta \times t\times \frac { 1 }{ tan\theta } =st\) 
\(h=\sqrt { st } \) 
hence, the height of tower is \(\sqrt { st } \)

Let the level of water raise in the tank in x hours = 15000x metres
Length of the water flow in x hours = 15000x metres
 Diameter of the pipe = 14 cm
radius = \(\frac { 14 }{ 2 } =7cm=\frac { 7 }{ 100 } cm\)
Volume of water = \(\pi r^{ 2 }h\)
Volume of water flow in x hours in the pond  =  \(\frac { 22 }{ 7 } \times \frac { 7 }{ 100 } \times \frac { 7 }{ 100 } \times 15000x\)
According to the question 
\(50\times 44\times \frac { 21 }{ 100 } =\frac { 22 }{ 7 } \times \frac { 7 }{ 100 } \times \frac { 7 }{ 100 } \times 15000x\)
\(\Rightarrow 22\times 21=\frac { 154\times 15 }{ 10 } x\Rightarrow x=\frac { 22\times 21\times 10 }{ 154\times 15 } =2\)
Hence, the level of water in the pond will rise by 21 cm in 2 hours. 

Let A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0) be the given points.
Then, \(\begin{aligned} A B= \sqrt{(1+1)^2+(0+2)^2}=\sqrt{(2)^2+(2)^2} \end{aligned}\)
\(\begin{aligned} {\left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right.} \end{aligned}\)
\(\begin{aligned} =\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \text { units } \\ \end{aligned}\)
\(\begin{aligned} B C= \sqrt{(-1-1)^2+(2-0)^2}=\sqrt{(-2)^2+(2)^2} \\ \end{aligned}\)
\(\begin{aligned} =\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \text { units } \\ \end{aligned}\)
\(\begin{aligned} C D=\sqrt{(-3+1)^2+(0-2)^2} \\ \end{aligned}\)
\(\begin{aligned} =\sqrt{(-2)^2+(-2)^2} \end{aligned}\)
\(\begin{aligned} =\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \text { units } \end{aligned}\)
\(\begin{aligned} D A =\sqrt{(-1+3)^2+(-2-0)^2} \end{aligned}\)
\(\begin{aligned} =\sqrt{(2)^2+(-2)^2}=\sqrt{4+4} \end{aligned}\)
\(\begin{aligned} =\sqrt{8}=2 \sqrt{2} \text { units } \end{aligned}\)
\(\begin{aligned} A C =\sqrt{(-1+1)^2+(2+2)^2} \end{aligned}\)
\(\begin{aligned} =\sqrt{0+4^2}=4 \text { units } \end{aligned}\)
and \(\begin{aligned} BD=\sqrt{(-3-1)^2+(0-0)^2} \end{aligned}\)
\(\begin{aligned} =\sqrt{(-4)^2+0}=4 \text { units } \end{aligned}\)
Here, the four sides AB, BC, CD and DA are equal and also diagonals AC and BD are equal.
So, the quadrilateral ABCD is a square.