Let AB be the height of the tip of the fishing rod from the water surface. Let BC be the horizontal distance of the fly from the tip of the fishing rod.
Then, AC is the length of the string.
AC can be found by applying Pythagoras theorem in ΔABC.
AC² = AB² + BC²
AB² = (1.8 m)² + (2.4 m)²
AB² = (3.24 + 5.76) m²
AB² = 9.00 m²
\(\Rightarrow A B=\sqrt{9} m=3 m\)
Thus, the length of the string out is 3 m.
She pulls the string at the rate of 5 cm per second.
Therefore, string pulled in 12 seconds = 12 x 5 = 60 cm = 0.6 m
Let the fly be at point D after 12 seconds.
Length of string out after 12 seconds is AD.
AD = AC - String pulled by Nazima in 12 seconds
= (3.00 - 0.6) m
= 2.4 m
In ΔADB,
AB² + BD² = AD^2­
(1.8 m)² + BD² = (2.4 m)²
BD² = (5.76 − 3.24) m² = 2.52 m²
BD = 1.587 m
Horizontal distance of fly = BD + 1.2 m
= (1.587 + 1.2) m
= 2.787 m
= 2.79 m

\(\frac { 1 }{ \sqrt { 3 } } \) 

0

A + B - C = 45⁰, B + C - A = 60⁰, A + B + C = 180⁰.
\(A={ 60 }^{ 0 },\quad B=\frac { { 105 }^{ 0 } }{ 2 } ,C=\frac { { 135 }^{ 0 } }{ 2 } \) 

A = 60⁰, B = 30⁰

\(\sin { A } =\frac { 24 }{ 25 } \)
\(\tan { C } =\frac { 7 }{ 24 } ;\cos { A } =\frac { 7 }{ 25 } \)
\(\sec { C } =\frac { 25 }{ 24 } \) 

(i)There are 500 possible ways to draw a card.Perfect squares of even numbers are 4, 16, 36, 64, 64, 100, 144,196, 256, 324, 400, 484.
Number of ways to draw a no. of which is a perfect square of even number =11
Probability of first player winning a prize = \(11\over 500\)
(ii)For second players No. of cards left = 500-1=499
[∵ One card has been drawn by 1st players and it has not been replaced]
Also No. of cards bearing perfect square of even number = 11
[∵ 1st players has not won ∴ Card drawn by him does not bear a perfect square even numbert]
∴ Probability of IInd player winning a prize = \(11\over 499\)

Total cards=20
(i)Probability of drawing a card bearing an odd number = \({10\over 20}={1\over 2}\)
(ii)Probability of drawing a card bearing an even number = \({10\over 20}={1\over 2}\)
(iii)Probability of drawing a prime number {2, 3, 5, 7, 11, 13, 17, 19}= \({8\over 20}={2\over 5}\)
(iv)Probability of drawing a number divisible by 3, {3, 6, 9, 12, 15, 18} = \({6\over 20}={3\over 10}\)
(v)Probability of drawing a number divisible 2 and 3 both (6, 12, 18) = \(3\over 20\)

Total outcomes
(H, H), (H, T), (T, 1), (T, 3), (T, 4), (T, 5), (T, 6)
A=(H, H), (H, T)
B=(T, 2), (T, 4), (T, 6)
C=(H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)
D=(T, 1), (T, 3), (T, 5)

Here, a = 8, l = 350, d = 9.
Using formula, l = a + ( n - 1 )d, we get
a + ( n - 1 )d = 350
\(\Rightarrow\)        8 + ( n - 1 )9 = 350   \(\Rightarrow\)   ( n - 1 )9 = 350 - 8
\(\Rightarrow\)    ( n - 1 )9 = 342   \(\Rightarrow\)    n - 1 = \({342 \over 9}\)
\(\Rightarrow\)    n - 1 = 38  \(\Rightarrow\) n = 38 + 1 = 39
From formula,       S_n = \({n\over 2}\) ( a + l ), we get
S₃₉ = \({39\over2}\) ( 8 + 350 ) = \({39 \over 2}\) X = 358 = 6981

Given, AP is \(\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48} \ldots\)
Here, a₁ \(=\sqrt{3}, a_{2}=\sqrt{12}=2 \sqrt{3}, a_{3}=\sqrt{27}=3 \sqrt{3}\)
and a₄ \(=\sqrt{48}=4 \sqrt{3} \) and so on.
\(\therefore\) Common difference (d) = a₂ - a₁
\(=2 \sqrt{3}-\sqrt{3}=\sqrt{3}\) 

From the given polynomial we will find the value, the sum of the zeroes, and the multiple of the zeroes.
\(\alpha+\beta=\frac{-b}{a}=\frac{-8}{1}=-8\)
\(\alpha\times\beta=\frac{c}{a}=\frac{6}{1}=6\)
Sum of zeroes = \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { -8 }{ 6 } =\frac { -4 }{ 3 } \)
Product of zeroes = \(\frac { 1 }{ \alpha } \times \frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =\frac { 1 }{ 6 } \)
Now for making a polynomial 
p(x) = x² - \((\alpha+\beta)x+\alpha\beta\)
\(\therefore\)  The polynomial is : \(p(x)=\frac { 1 }{ 6 } \left( 6{ x }^{ 2 }+8x+1 \right) \)

Given, p(x) = 3x² - 4x - 7 and \(\alpha\) and \(\beta\) are zeroes.
Sum of zeroes = \(\alpha +\beta =-\frac { Coefficient \ of \ x }{ Coefficient \ of \ { x }^{ 2 } } \)
\(=\left( -\frac { 4 }{ 3 } \right) =\frac { 4 }{ 3 } \)
Product of zeroes = \(\alpha \beta =\frac { Constant \ term }{ Coefficient \ of \ { x }^{ 2 } } \)
\(=\left( \frac { -7 }{ 3 } \right) =\frac { 7 }{ 3 } \)
For the new polynomial,
Sum of zeroes = \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { \frac { 4 }{ 3 } }{ -\frac { 7 }{ 3 } } =\frac { -4 }{ 7 } \)
Product of zeroes = \(\frac { 1 }{ \alpha } \times \frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =\frac { 1 }{ -\frac { 7 }{ 3 } } =\frac { -3 }{ 7 } \)
\(\therefore\)  Required quadratic polynomial = x² - (Sum of zeroes)x + Product of zeroes
\({ x }^{ 2 }-\left( \frac { -4 }{ 7 } \right) x+\left( \frac { -3 }{ 7 } \right) \)
\(=\frac { 1 }{ 7 } \left( 7{ x }^{ 2 }+4x-3 \right) \)
\( =\left( 7{ x }^{ 2 }+4x-3 \right) \frac { 1 }{ 7 }\)

Given, polynomial, f(x) = ax² - 5x + c
Let the zeroes of f(x) are \(\alpha\) and \(\beta\), then according to the question
Sum of zeroes, \((\alpha+\beta)\) = Product of zeroes, \((\alpha\beta)\) = 10
Now \(\alpha +\beta =-\frac { Coeff.of \ x }{ Coeff \ of \ { x }^{ 2 } } =\frac { -5 }{ a } \)
\(\Rightarrow \quad 10=\frac{+5}{a}\)
\(\therefore \quad a=\frac{1}{2}\)
and \(\alpha \beta =\frac { Constant \ term }{ Coeff \ of \ { x }^{ 2 } } \)
\(\Rightarrow \quad 10 = 2c\)
\(\therefore \quad c=5\)
Hence \(a=\frac{1}{2}\) and c = 5

(i) Now p(x) = 2x³ - 11x² + 17x - 6
p(2) = 2(2)³ - 11(2)² + 17(2) - 6
= 16 - 44 + 34 - 6
= 50 - 50
= 0
Hence 2 is the zero of p(x)
(ii) Again  p(3) = 2(3)³ - 11(3)² + 17(3) - 6
= 54 - 99 + 51 - 6
= 105 - 105 = 0
Hence, 3 is zero of p(x)
(iii) Again  \(p\left( \frac { 1 }{ 2 } \right) =2\left( \frac { 1 }{ 2 } \right) ^{ 3 }-11\left( \frac { 1 }{ 2 } \right) ^{ 2 }+17\left( \frac { 1 }{ 2 } \right) -6\)
\(=\frac { 1 }{ 4 } -\frac { 11 }{ 4 } +\frac { 17 }{ 2 } -6\)
= 0
Hence, \(\frac{1}{2}\) is also the zero of p(x).

Let    x=\(0.\overline { 3178 } \)
⇒ x =.3178178178...
⇒ 10,000 x = 3178.178178
⇒ 10 x =3.178178...
substracting,  9990 x = 3175
⇒ \(x=\frac { 3175 }{ 9990 } =\frac { 635 }{ 1998 } \)

\(\frac { 15 }{ 4 } +\frac { 5 }{ 40 } =\frac { 15 }{ 4 } \times \frac { 25 }{ 25 } +\frac { 5 }{ 40 } \times \frac { 25 }{ 25 } \)
​​​​\(=\frac { 375 }{ 100 } +\frac { 125 }{ 1000 } \)
= 3.75 + 0.125 = 3.875

 We have, \(\frac{3}{8}\)
The factors of the denominator 8 is 2³ x 5⁰.
So, \(\frac{3}{8}\) has a terminating decimal expansion.
Now, \(\frac { 3 }{ 8 } =\frac { 3 }{ { 2 }^{ 3 } } \times \frac { { 5 }^{ 3 } }{ { 5 }^{ 3 } } =\frac { 375 }{ { 10 }^{ 3 } } =0.375\)
which is the required decimal expansion.

Let the vertices of △ABC be A(x₁,y₁),  B(x₂,y₂) and C(x₃, y₃). Let D(3, 1), F.(5, 6) and F(- 3, 2) be the mid-points of BC, CA and AB respectively. 

then \(\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =3\ \Rightarrow \ { x }_{ 2 }+{ x }_{ 3 }=6\) ---- (i)
\(\frac { y_{ 2 }+{ y }_{ 3 } }{ 2 } =3\ \Rightarrow \ { y }_{ 2 }+{ y }_{ 3 }=2\) --- (ii)
\(\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =5\ \Rightarrow \ { x }_{ 3 }+{ x }_{ 1 }=10\) --- (iii) and \(\frac { y_{ 3 }+{ y }_{ 1 } }{ 2 } =6\ \Rightarrow \ { y }_{ 1 }+{ y }_{ 3 }=12\) --- (iv)
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =-3\ \Rightarrow \ { x }_{ 1 }+{ x }_{ 2 }=-6\) --- (v) and \(\frac { y_{ 1 }+{ y }_{ 2 } }{ 2 } =2\ \Rightarrow \ { y }_{ 1 }+{ y }_{ 2 }=4\) --- (vi)
Adding (i), (iii) and (v), we get
2(x₁+x₂+x₃) = 10 ⇒ x₁+x₂+x₃=5 --- (vii)
Subtrcting (i), (iii), (v) separtely from (vii), we get x₁=-1 , x₂=-5, x₃=11 
Adding (ii), (iv) and (vi), we get
2(y₁+y₂+y₃) = 18 ⇒ y₁+y₂+y₃=9  --- (viii)
Subtracting (ii), (iv) and (vi) separetely from (viii), we get y₁=7, y₂=-3, y₃=5.
Hence, the vertices of the triangle ABC are A(-1,7), B(-5,-3) and C(11,5).

Let P(x,y), A(a+b, a-b) and B(a-b, a+b) be the given points.
Since AP = BP,
ஃ AP² = BP²
⇒ (x-a-b)²+(y-b+a)² = (x-a+b)²+(y-a-b)²
⇒ (x-a-b)²- (x-a+b)² = (y-a-b)²-(y-b+a)²
⇒ (x-a-b+x-a+b)(x-a-b-x+a-b) = (y-a-b+y-b+a) (y-a-b-y+b-a)
⇒ (2x-2a)(-2b) = (2y-2b)(-2a) ⇒ -4bx+4ab = -4ay+4ab
⇒ -4bx = -4ay ⇒ bx = ay. Hence proved

AB = \(\sqrt { { (3-0 })^{ 2 }+(\sqrt { 3 } -0)^{ 2 } } \) = \(\sqrt{9+3}\) =\(\sqrt {12}\) = 2\(\sqrt{3}\)
∴ Side of equilateral △ = 2\(\sqrt{3}\) units
Area of equilateral △ = \(\frac{\sqrt 3} {4}\) (Sides)²
=\(\frac {\sqrt {3}} {4} \times (2\sqrt {3})^{2} = 3{\sqrt{3}}\) sq.units