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Let P be the position of the aeroplane. Then, PM=200m and let A and B be two points on the two banks of a river such that the angles of depression at A and B are \({ 60 }^{ \circ }\)and \({ 45 }^{ \circ }\) , respectively. 
Let  Am=x m and BM= y m.
Then, \(\angle XPB=\angle MBP={ 45 }^{ \circ }\)  [altenatives angles]
and  \(\angle XPB=\angle MBP={ 60 }^{ \circ }\)  [altenatives angles]
In right angled  \(\Delta AMP,\) \(\tan { { 60 }^{ \circ } } =\frac { PM }{ AM } \)
\(\Rightarrow \sqrt { 3 } =\frac { 200 }{ x } \Rightarrow 200=\sqrt { 3x } \)
\(\Rightarrow x=\frac { 200 }{ \sqrt { 3 } } m \)
In right angled  \(\Delta BMP,\)
\(\tan { { 45 }^{ \circ } } =\frac { PM }{ BM } \Rightarrow 1=\frac { 200 }{ y }\)
\(\Rightarrow y=200m \)
Now, width of the river, AB=BM+MA
\(\Rightarrow AB=x+y=\frac { 200 }{ \sqrt { 3 } } +200\)
\( =200\left( \frac { 1 }{ \sqrt { 3 } } +1 \right)\)
\(=200(1.5773)=315.46m. [\because \sqrt { 3 } =1.732]\)
Hence, the width of the river is 315.46m.

P(-1,-1)

Let AO = OB = r
Then perimeter of semicircle APB
= \(2\pi r \over 2\) = \(\pi\)r
Perimeter of semicircle AQO
= \(\frac { 2\pi \frac { r }{ 2 } }{ 2 } =\frac { \pi r }{ 2 } \)
(∵ radius of semicircle AQO = \(r\over2\))
∴ Perimeter of shaded region
\(=\ \pi r+\frac { \pi r }{ 2 } +r=\frac { 2\pi r+\pi r+2r }{ 2 } \)
But perimeter of shaded region
ஃ \(\frac { 2\pi r+\pi r+2r }{ 2 } =40\)
⇒ r(2\(\pi\) + \(\pi\) + 2) = 80
⇒  r(3\(\pi\) + 2) = 80
⇒ \(r\left( 3\times \frac { 22 }{ 7 } +2 \right) =80\)
⇒ \(r\left( \frac { 66 }{ 7 } +2 \right) =80\)
⇒  \(r\left( \frac { 80 }{ 7 } \right) =80\)
ஃ  \(r=\frac { 80\times 7 }{ 80 } =7cm\)
Now, Area of APB = \(\frac { \pi { r }^{ 2 } }{ 2 } =\frac { 22\times 7\times 7 }{ 7\times 2 } \)
= 77 cm2
Area of AQO = \(\frac { \pi { r }^{ 2 } }{ 2 } \)
 \(=\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times \frac { 1 }{ 2 } \)
= \(77\over4\) cm²
ஃ Area of shaded region
= 77 + \(\frac { 77 }{ 4 } =\frac { 308+77 }{ 4 } \)
= \(\frac { 385 }{ 4 } \) = 96.25 cm²

Let Ist term =  a, common difference = d.
\(\therefore\)  a_m = a + ( m - 1 )d and a_n = a + ( n - 1 )d
A.T.Q.,
m.a_m = n.a_n                      \(\Rightarrow\) m { a + ( m - 1 )d} = n{ a + ( n - 1 )d }
\(\Rightarrow\)    ma - na = n( n -1 )d - m ( m - 1)d - m( m - 1 )d   \(\Rightarrow\) ( m - n ) a = ( n² - n - m² + m )d
\(\Rightarrow\)   ( m - n ) a = ( n - m ) ( m + n - 1 )d   \(\Rightarrow\) a = - ( m + n - 1 )d
Now    a_{m + n} = a + ( m + n - 1)d = - ( m + n - 1 )d + ( m + n - 1 )d = 0

Let C be the point where the ball is
ㄥC = 60° (alternate angles)
In ΔABC, \(tan\ 60^0={AB\over BC}\)
⇒ \(\sqrt3={20\over x}\)
⇒ \(x={20\over \sqrt3}\)
\(=20\left(\sqrt3\over3 \right)\)
Hence, required distance is
1 = 11.53m

Volume of cylinder = Volume of sphere, 
\(\pi\)r²h =\(\pi\)
where r and h are radius of base and height of cylinder 
(0.5)²h=1
\({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }h=1\)
h=4 cm.

Three consecutive terms k + 2, 4k - 6 and 3k - 2 are in A.P.
\(\therefore\) (4k-6)-(k+2) = (3k-2)-(4k-6)
\(\Rightarrow\) 4k - 6 - k - 2 = 3k - 2 - 4k + 6
\(\Rightarrow\) 3k - 8 = - k + 4
\(\Rightarrow\) 4k=4+8
\(\Rightarrow\) k = 12/4 =3

p=-5, q=-6

Given, k -1, k + 3 and 3k - 1 are in AP.
\(\therefore\)  Second term - First term = Third term - Second term
\(\Rightarrow (k+3)-(k-1)=(3k-1)-(k+3)\) 
\(\Rightarrow\) 4 = 2k - 4
\(\Rightarrow\)  k = 4

\(\frac { 1 }{ 5 } \)

Here, speed of the motor cycle = 66km/h
\(={66000\over60}m/min\)
=1100 m/min
Radius of the wheel \((r)=35cm={35\over 100}m\)
circumference of the wheel=\(2\times{22\over 7}\times{35\over 100}m\)
=2.2m
Number of revolutions per minute=\({1100\over 2.2}\)
=500.

4 cm

66.67m

8 cm

Here, \(|AB|=\sqrt { \left( 1+3 \right) ^{ 2 }+\left( -3-0 \right) ^{ 2 } } \)
\(=\sqrt { { 4 }^{ 2 }+(-3)^{ 2 } } =\sqrt { 16+9 } \)
\(=\sqrt { 25 } =5\quad units\)
\(|BC|=\sqrt { \left( 4-1 \right) ^{ 2 }+\left( 1+3 \right) ^{ 2 } } =\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } \)
\(=\sqrt { 9+16 } =\sqrt { 25 } =5\quad units\)
\(|CA|=\sqrt { \left( -3-4 \right) ^{ 2 }+\left( 0-1 \right) ^{ 2 } } \)
\(=\sqrt { \left( -7 \right) ^{ 2 }+\left( -1 \right) ^{ 2 } } \)
\(=\sqrt { 49+1 } =\sqrt { 50 } =5\sqrt { 2 } units\)
Now, AB = BC and AB² + BC² CA² 
Thus, ΔABC is an isosceles right triangle.

Two-digit odd positive numbers are 11,
13, 15, ... 99.
These numbers are in AP.
Here    a =  11 and d = 2
a_n = 99
\(\Rightarrow\) a + ( n - 1 )d = 99
\(\Rightarrow\) 11 + ( n - 1 ) x 2 = 99
\(\Rightarrow\) ( n - 1 ) x 2 = 88
\(\Rightarrow\)   n - 1 = 44 \(\Rightarrow\)  n = 45
Since,   S_n = \(\frac{n}{2}(a+l)\)
\(\Rightarrow\)   S₄₅ = \(\frac{45}{2}(11+99)=2475\)

Part of tank fillled by one pipe in 1 hour - PArt of tank emptied by IInd pipe in 1 hour = part of tank filled by both the pipes in 1 hour
\(\Rightarrow \frac { 1 }{ x-2 } -\frac { 1 }{ x+2 } =\frac { 1 }{ 24 } \)
\(\Rightarrow \frac { x+2-x+2 }{ (x-2)(x+2) } =\frac { 1 }{ 24 } \)
\(\Rightarrow x^{ 2 }-4=96\)
\(\Rightarrow x^{ 2 }=100\)
\(\Rightarrow \) x=10, -10 [Rejected x=-10]
\(\therefore \) Time taken by the pipe to empty the tank = 10+2=12 hours

It can be observed from the given data that the maximum class frequency is 10 belonging to class interval 30 − 35.
Therefore, modal class = 30 − 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f₁) of modal class = 10
Frequency (f₀) of class preceding modal class = 9
Frequency (f₂) of class succeeding modal class = 3
\(\text { Mode }=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}} \times h\right) \)
\(=30+\left(\frac{10-9}{2}(10)-9-3\right) \times(5) \)
\(=30+\left(\frac{1}{20-12}\right) 5\)
= 30 + 5/8 = 30.625
Mode = 30.6
It represents that most of the states/U.T have a teacher-student ratio as 30.6.
To find the class marks, the following relation is used.
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Taking 32.5 as assumed mean (a), d_i, u_i, and f_iu_i are calculated as follows.

\(\text { Mean, } \bar{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) \times h \)
\(=32.5+\left(\frac{-23}{35}\right) \times 5\)

= 32.5 - 23/7 = 32.5 - 3.28
= 29.22
Therefore, mean of the data is 29.2.
It represents that on an average, teacher−student ratio was 29.2

LHS =  \(\frac { \frac { 1 }{ \cos ^{ 2 }{ \theta } } -\sin ^{ 2 }{ \theta } }{ \frac { \sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } } =\frac { 1-\sin ^{ 2 }{ \theta } .\cos ^{ 2 }{ \theta } }{ \sin ^{ 2 }{ \theta } } ={ cosec }^{ 2 }\theta -\cos ^{ 2 }{ \theta } \)
\(=1+\cot ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } \) 

Given, pair of linear equations is x - y + 1 = 0 and 3x + 2y -12 = O. Table for x r- y+l = 0 or y = x + 1 is

Table for 3x+2y-12=0 or \(y=\frac {12-3x}{2}\) is

Now, plot the points A (0,1), B(4, 5), C (-1,0) and join them to get a line CB. Similarly, plot the points D(0, 6), E(2,3), F (4,0) and join them to get a line DF. 

Clearly, the two lines intersect each other at the point E (2, 3). Hence, x = 2 and y = 3 is the solution of the given pair of equations. The line DE cuts X-axis at the point F (4, 0) and the line AB cuts X-axis at the point C (-1, 0).
Hence, the coordinates of the vertices of the triangle, so formed are E(2, 3), F( 4,0) and C (-1,0).

p=2, q=3

Given, \(a+\sqrt { b } =c+\sqrt { d } \)
If a = c, then \(\sqrt { b } =\sqrt { d } \)
\(\Rightarrow \) b = d      [squaring both sides]
If \(a\neq c\), then there exists a positive rational number x such that a = c + x.
Now, \(a+\sqrt { b } =c+\sqrt { d } \)
\(\Rightarrow \quad c+x+\sqrt { b } =c+\sqrt { d } \quad \quad \left[ \because \quad a=c+x \right] \)
\(\Rightarrow \quad x+\sqrt { b } =\sqrt { d } \)
\(\Rightarrow \quad \left( x+\sqrt { b } \right) ^{ 2 }=\left( \sqrt { d } \right) ^{ 2 }\)    [squaring both sides]
\(\Rightarrow \quad { x }^{ 2 }+b+2x\sqrt { b } =d\quad \Rightarrow { x }^{ 2 }+2x\sqrt { b } +b-d=0\)
\(\Rightarrow \quad 2x\sqrt { b } =d-b-{ x }^{ 2 }\Rightarrow \sqrt { b } =\frac { d-b-{ x }^{ 2 } }{ 2x } \)
Since, d, x and b are rational numbers and x > 0.
So, \(\frac { d-b-{ x }^{ 2 } }{ 2x } \) is a rational number.
\(\Rightarrow \) \(\sqrt { b } \) is a ratonal number.
\(\Rightarrow \)  b is the square of a rational number.
from Eq.(i)
\(\sqrt { d } =x+\sqrt { b } \quad \Rightarrow \quad \sqrt { d } \) is a rational number.
\(\Rightarrow \) d is the square of a rational number.
Hence, either a = c and b = d or b and d are the squares of rationals.

(75.36 - \(36\sqrt { 3 } \)) cm²

There are 40 students, and only one name card has to be chosen.
(i) The number of all possible outcomes is 40
The number of outcomes favourable for a card with the name of a girl = 25  
Therefore, P (card with name of a girl) = P(Girl) = \(\frac{25}{40}=\frac{5}{8}\)
(ii) The number of outcomes favourable for a card with the name of a boy = 15 
Therefore, P(card with name of a boy) = P(Boy) \(=\frac{15}{40}=\frac{3}{8}\)
Note : We can also determine P(Boy), by taking
P(Boy) = 1 – P(not Boy) = 1 – P(Girl) \(=1-\frac{5}{8}=\frac{3}{8}\)

R₁ = \(\frac { 40 }{ 2 } \) cm , R₂ = \(\frac { 20 }{ 2 } \)= 10 cm,height = 12cm 
Volume = \(\frac { 1 }{ 3 } \) \(\pi h\) (R²₁ + R²₂  +R₁R₂) 
\(\frac { 1 }{ 3 } \times 12\left( 20^{ 2 }+10^{ 2 }+20\times 10 \right) \)
= \(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 12\times 700=8800cm^{ 3 }\)
Slant height t = \(\sqrt { h^{ 2 }+(R_{ 1 }-R_{ 2 })^{ 2 } } \)
\(\sqrt { (12)^{ 2 }+(20-10)^{ 2 } } =\sqrt { 244\quad } cm=15.62cm\)
Area of tin sheet used = \(\pi R^{ 2 }_{ 2 }+\pi l(R_{ 1 }+R_{ 2 })\)
= \(\pi \left[ R^{ 1 }_{ 2 }+l(R_{ 1 }+R_{ 2 }) \right] =\frac { 22 }{ 7 } \left( 10 \right) ^{ 2 }+15.62\times 30=1787.02cm^{ 3 }\)
Cost of 1 dm² of sheet used = Rs 1.20
\(\Rightarrow \) cost 100 cm² of sheet = Rs 1.20
\(\Rightarrow \) Cost of 1787.02 cm² sheet = Rs \(\frac { 1.20 }{ 100 } \times 1787.02=\) Rs 21.44

Let cloud is at C and its reflection is C'. B is a point h m , above lake such that ㄥCBE=α and ㄥEBC'=β
In right ∆BEC', \(\frac { EC }{ BE } \)=tanβ
In right ΔCEB. \(\frac { CE }{ BE } =tan\alpha ,\frac { BC }{ BE } =sec\alpha \)
Now \(\frac { 2h\quad sec\alpha }{ tan\beta -tan\alpha } =\frac { 2h\times \frac { BC }{ BE } }{ \frac { EC' }{ BE } -\frac { CE }{ BE } } \)
=\(\frac { 2h\times BC }{ EC'-CE } =\frac { 2h\times BC }{ (h+x+h)-x } =\frac { 2h\times BC }{ 2h } \)=BC
BC=distance of the cloud from the point of observation. Hence proved.

Let points P and Q trisect the line joining the points.

∴ AP=PQ=QB
P divides AB in the ration 1:2 and Q divides AB in the ration 2:1
P (coordinate x) = \(\frac { 1\times \left( -2 \right) +2\times 4 }{ 1+2 } =\frac { 6 }{ 3 } \)=2; P (coordinate y) = \(\frac { 1\times \left( -3 \right) +2\times \left( -1 \right) }{ 1+2 } =-\frac { 5 }{ 3 } \)
The coordinates of P are \(\left( 2,\frac { 5 }{ 3 } \right) \)
Q(x-coordinate) =  \(\frac { 2\times \left( -2 \right) +1\times \left( 4 \right) }{ 2+1 } =-\frac { -4+4 }{ 3 } =0\) Q(y-coordinates) = \(\frac { 2\times \left( -3 \right) +1\times \left( -1 \right) }{ 2+1 } =-\frac { -7 }{ 3 } \)
The coordinates of Q are \(\left( 0,-\frac { 7 }{ 3 } \right) \).