sin 3 (10°) sin 6 (10°) sin 9 (10°)
= \(\cfrac { 1 }{ 2 } \times \cfrac { \sqrt { 3 } }{ 2 } \times 1=\cfrac { \sqrt { 3 } }{4 } \)

\(cot\theta =1\)
\(cot\theta =\cfrac { 1 }{ 3 } \)
\(\cfrac { adjacent }{ opposite } =\cfrac { 1 }{ 3 } \)
\(\sqrt { { 3 }^{ 2 }+1 } =\sqrt { 10 } \)
\(\cfrac { 3cos\theta -4sin\theta }{ 5sin\theta +4cos\theta } =\cfrac { 3\times \cfrac { 1 }{ \sqrt { 10 } } -4\times \cfrac { 3 }{ \sqrt { 10 } } }{ 5\times \cfrac { 3 }{ \sqrt { 10 } } +4\times \cfrac { 1 }{ \sqrt { 10 } } } =\cfrac { 3-12 }{ 15+4 } =\cfrac { -9 }{ 19 } \)

3a = 150
a = \(\frac { 150 }{ 3 } \) = 50 m

\((9,-3)=\left( \frac { m(12+n(7) }{ m+n } ,\frac { m(1)+n(-5) }{ m+n } \right) \)
\(\frac {13m+37n}{ m+n } =9\)
13m + 7n = 9m + 9n
4m = 2n
\(\frac { m }{ n } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)
\(\frac{m-5n}{m+nn}=-3\)
m -5n = -3m -3n
m + 3m= 5n-3n
4m=2n
\(\frac { m }{ n } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)

\((5,1)=\left( \frac { { x }_{ 1 }-3 }{ 2 } ,\frac { { y }_{ 1 }+5 }{ 2 } \right) \)
\(\frac { { x }_{ 1 }-3 }{ 2 } =5,\quad \frac { { y }_{ 1 }+5 }{ 2 } \)
x₁- 3 = 10, y₁+ 5 = 2
x₁ = 13,  y₁ = -3
B(13, -3)
\((-5,-1)=\left( \frac { -3+{ x }_{ 2 } }{ 2 } ,\frac { 5+{ y }_{ 2 } }{ 2 } \right) \)
\(\frac { -3+{ x }_{ 2 } }{ 2 } -5,\)
-3 + x₂ = - 10
X₂ = -7,
C (-7, -7)
\(\frac { 5+{ y }_{ 2 } }{ 2 } =-1\)
5+ y₂ = -2
y₂ = -7
\(\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( \frac { 13+(-7) }{ 2 } ,\frac { -3+(-7) }{ 2 } \right) =(3,-5)\)

Total no. of attempts n(S) = 40
Total no. of attempts by A team n(A) = 32
Total no. of attempts by the opponent team B = n(B) = 40 - 32 = 8
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 8 }{ 40 } =\frac { 1 }{ 5 } \).

x₁ y₁ x₂ y₂
A(-2, -3) B(2, 1)
m : n = 3 : 1
The point P divides AB in the ratio 3 : 1
\(P(x,y)=\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\(=\left( \frac { 3\times 2+1\times -2 }{ 3+1 } ,\frac { 3\times 1+1\times -3) }{ 3+1 } \right) \)
\(=\left( \frac { 6-2 }{ 4 } ,\frac { 3-3 }{ 4 } \right) =\left( \frac { 4 }{ 4 } ,\frac { 0 }{ 4 } \right) =(1,0)\)
P is at 25% distance from B on its left and C is at 25% distance from B on its right
\(\therefore\) B is the mid point of PC
Mid point of  \(\bar { PC } =\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\((2,1)=\left( \frac { 1+{ x }_{ 2 } }{ 2 } ,\frac { 0+{ y }_{ 2 } }{ 2 } \right) \)
\(\frac { 1+{ x }_{ 2 } }{ 2 } =2\quad \frac { { y }_{ 2 } }{ 2 } =1\)
1 + x₂ = 4 y₂ = 2
⇒ x₂ = 3, y₂ = 2
∴ C(x₂, y₂) = (3, 2) is the solutions.

\(\sqrt [ 4 ]{ 5 } \) = \({ 5 }^{ \frac { 1 }{ 2 } }={ 5 }^{ \frac { 3 }{ 12 } }=\sqrt [ 12 ]{ { 5 }^{ 3 } } =\sqrt [ 12 ]{ 125 } \)

6 has the maximum frequency 9. Therefore 6 is the mode.

\(\sqrt [ 3 ]{ 40 } \times \sqrt [ 3 ]{ 16 } =\left( \sqrt [ 3 ]{ 2\times 2\times 2\times 5 } \right) \times \left( \sqrt [ 3 ]{ 2\times 2\times 2\times 2 } \right) \)
=\(\left( 2\times \sqrt [ 3 ]{ 5 } \right) \times \left( 2\times \sqrt [ 3 ]{ 2 } \right) =4\times \left( \sqrt [ 3 ]{ 2 } \times \sqrt [ 3 ]{ 5 } \right) =4\times \sqrt [ 3 ]{ 2\times 5 } \)
= \(4\sqrt [ 3 ]{ 10 } \) .

\(\frac{1}{5}\) = 5^-1

n(A) = 300, n(A∪B) = 500, n(A∩B) = 50 and n(B') = 350
n(A∪B) = n(A) + n(B) - n(A∩B)
500 = 300 + n(B) - 50
500 = 250 + n(B)
500 - 250 = n(B)
250 = n(B)
\(\therefore\) n(B) = 250
n(U) = n(B) + n(B)'
= 250 + 350
= 600
\(\therefore\) n(B) = 250 and n(U) = 600

f(x) = x² - 4x + 3

Since the value of the polynomial f(x) at x = 1 and x = 3 is zero, as the zeros of polynomial f(x) are 1 and 3.

Supplement of Right angle (90°) = 180° - 90°
= 90°

Let p (x) 2x³ - x² - 12x - 9
Sum of the co-efficients = 2 - 1- 12- 9 = -20 \(\neq \) 0
Hence x-1 is not a factor
Sum of co-efficients of even powers with constant = -1 - 9 = -10
Sum of co-efficients of odd powers = 2 - 12= -10
Hence x + 1 is a factor of x.
Now we use synthetic division to find the other factors.

Then p (x) = (x + 1)(2x² - 3x - 9)
Now 2x² - 3x - 9 = 2x² - 6x + 3x - 9 = 2x (x - 3) + 3 (x - 3)
= (x - 3)(2x + 3)
Hence 2x³ - x² - 12x - 9 (x + 1) (x - 3) (2x + 3)

p(x₁) = 2x³ + ax² + 4x - 12
When it is divided by x - 3,
p(3) = 2(3)³ + a(3)² + 4(3) - 12
= 54 + 9a + 12 - 12
= 54 + 9a .....(R₁)
p(x2) = x³ + x² - 2x + a
When it is divided by x - 3,
p(3) = 3³ + 3² - 2(3) + a
= 27 + 9 -6 +a
= 30 + a  ......(R₂)
The given remainders are same (R₁ = R₂)
\(\therefore\) 54 + 9a = 30 + a
9a - a = 30 - 54
8a = -24
\(\therefore\)  a = - 24/8 = - 3
Consider R₂,
Remainder = 30 - 3
= 27

Given: AB = 2, BC = 6, AE = 6, BF = 8, CE = 7 and CF = 7
Consider ΔAEC and ΔBCF.
In ΔAEC, AE = 6, EC = 7 and AC = 8 (2 + 6 = 8)
In ΔBCF, BC = 6, CF = 1 and BF = 8
∴ ΔAEC ≅ BCF
∴ Area of ΔAEC = Area of ΔBCF (Two triangles are similar areas are equal)
Subtract area of ΔBDC on both sides we get,
Area of ΔAEC - Area of ΔBDC = Area of ΔBCF - Area of ΔBDC
Area of quadrilateral ABDE = Area of ΔCDF.
The answer is 1 : 1.

Total number of staff members = 42
The relative frequencies:
Car users \(=\frac { 7 }{ 42 } =\frac { 1 }{ 6 } \)
Two-wheeler users \(=\frac { 20 }{ 42 } =\frac { 10 }{ 21 } \)
Cycle users \(=\frac { 15 }{ 42 } =\frac { 5 }{ 14 } \)

Here, Total Surface Area of the cube = 486 cm²
6a² = 486 \(\Rightarrow\) a² = \(\frac{486}{6}\) and so, a² = 81 . This gives a = 9.
The side of the cube = 9 cm
Lateral Surface Area = 4a² = 4 × 9² = 4 × 81 = 324 cm²

(i) We know that cosec A =  sec(90⁰ A)
sec(90⁰ - A) sec(34⁰)
90⁰ - A =34⁰
We get A = 90° − 34°
A = 56⁰
(ii) We know that tan B = cot(90⁰ - B)
cot(90⁰ - B) = cot 47⁰
90⁰ - B = 47⁰
We get B = 90° − 47°
B = 43⁰

(i) 8x- 3y = 12      ...(1)
5x-2y = 7      ..(2)
8x- 3y-12 = 0
5x-2y- 7 = 0
For cross multiplication method, we write the co-efficients as

\(\cfrac { x }{ (-3)(-7)(-2)(-12) } =\cfrac { y }{ (-12)(5)-(-7)(8) } =\cfrac { 1 }{ (8)(-2)-(5)(-3) } \)
\(\cfrac { x }{ 21-24 } =\cfrac { y }{ -60+56 } =\cfrac { 1 }{ -16+15 } \)
\(\cfrac { x }{ -3 } =\cfrac { y }{ -60+56 } =\cfrac { 1 }{ -16+15 } \)
\(\therefore \ \cfrac { x }{ 3 } =\cfrac { 1 }{ -1 } \ \cfrac { y }{ -4 } =\cfrac { 1 }{ -1 } \)
x = 3 , y = 4
\(\therefore\)  Solutions: x = 3; y = 4
(ii) 6x+ 7y-11 = 0
5x+ 2y-13 = 0
For cross multiplication method, we write the co-efficients as
\(\cfrac { x }{ -91-(-22) } =\cfrac { y }{ -55-(-68) } =\cfrac { 1 }{ 12-35 } \)
\(\cfrac { x }{ -91+22 } =\cfrac { y }{ -55+78 } =\cfrac { 1 }{ -23 } \)
\(\cfrac { x }{ -69 } =\cfrac { y }{ 23 } =\cfrac { 1 }{ -23 } \)
\(\cfrac { x }{ -69 } =\cfrac { 1 }{ -23 } \quad \cfrac { y }{ 23 } =\cfrac { 1 }{ -23 } \)

x = 3, y= -1
\(\therefore\)  x = 3; y= -1
(iii) 

\(\cfrac { 2 }{ x } +\cfrac { 3 }{ y } -5=0\)
\(\cfrac { 3 }{ x } -\cfrac { 1 }{ y } +9=0\)
In (1), (2) Put \(\cfrac { 1 }{ x } =a,\cfrac { 1 }{ y } =b\)
(1) \(\Rightarrow\) 2a + 3b - 5 = 0
(2) \(\Rightarrow\) 3a - b + 9 = 0
For cross multiplication method, we write the co-efficients as
\(\cfrac { a }{ (3)(9)-(-1)(-5) } =\cfrac { b }{ (-5)(3)-(9)(2) } =\cfrac { 1 }{ (2)(-1)-(3)(3) } \)
\(\cfrac { a }{ 27-5 } =\cfrac { b }{ -15-18 } =\cfrac { 1 }{ -2-9 } \)
\(\cfrac { a }{ 22 } =\cfrac { b }{ -33 } =\cfrac { 1 }{ -11 } \)
\(\therefore \ \cfrac { a }{ 22 } =\cfrac { 1 }{ -11 } \cfrac { b }{ -33 } =\cfrac { 1 }{ -11 } \)

a = -2    b = 3
\(a=\cfrac { 1 }{ x } =-2\quad b=\cfrac { 1 }{ y } =3\)
\(\therefore \ x=\cfrac { -1 }{ 2 } \ \ y=\cfrac { 1 }{ 3 } \)
solution \(\ x=\cfrac { -1 }{ 2 } \ \ y=\cfrac { 1 }{ 3 } \)

Let POQ be the right angled triangle and O be placed at the origin. Let OQ = a units and OP be b units. Let us name the coordinates of P as (0,b) and Q as (a,0).
By mid-point formula, if M is the mid-point of the hypotenuse PQ [PM=MQ], then M is

\(\left( \frac { a+0 }{ 2 } ,\frac { b+0 }{ 2 } \right) =\left( \frac { a }{ 2 } ,\frac { b }{ 2 } \right) \)
We now use the distance formula and find that
OM=\(\sqrt { { \left( \frac { a }{ 2 } -0 \right) }^{ 2 }{ +\left( \frac { b }{ 2 } -0 \right) }^{ 2 } } =\sqrt { \frac { { a }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } } \) which is the same value as
QM=\(\sqrt { { \left( a-\frac { a }{ 2 } \right) }^{ 2 }{ +\left( 0-\frac { b }{ 2 } \right) }^{ 2 } } =\sqrt { \frac { { a }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } } \) and similarly PM=\(\sqrt { \frac { { a }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } } \)
This shows OM = QM = PM, which we desired to prove.

\(A\cap B\) = {5,7}
\(B\cap A\)= {5,7}
From (3) and (4) we get, \(A\cap B=B\cap C\)
It is verifield that insersection of sets is commutative.

Let p(x)= x³ + 9x² - x - 105
By factor theorem,x-3 is a factor of p(x),if p(3) = 0
p(3) = 33 + 9(3)³ - 3 - 105
= 27 + 81 - 3 -105
= 108-108
p(3) = 0
Therefore,x-3 is a factor of x³ + 9x² - x - 105

\(\sqrt { 44 } +\sqrt { 99 } -\sqrt { 275 } \) = \(\sqrt { 4\times 11 } +\sqrt { 9\times 11 } -\sqrt { 11\times 25 } \)
= \(\left( 2\sqrt { 11 } +3\sqrt { 11 } \right) -5\sqrt { 11 } =5\sqrt { 11 } -5\sqrt { 11 } =0\)

\(\bar { x } =20.2\) 
\( \bar{x}=\cfrac{\Sigma fx}{\Sigma f}\)
\(=\cfrac { 10\times 6+15\times 8+20p+25\times 10+30\times 6 }{ 6+8+p+10+6 } \)
\(20.2=\cfrac { 60+120+20p+250+180 }{ 30+p } \)
(30+p)20.2 = 610+20 p
606+20.2p = 610+20 p
20.2 p-20 p = 610-606 = 4
0.2 p = 4
\(\Rightarrow \) \(p=\cfrac { 4\times 10 }{ 0.2\times 10 } =\cfrac { 40 }{ 2 } =20\)

\(\frac { \sqrt { 5 } }{ \sqrt { 6 } +2 } -\frac { \sqrt { 5 } }{ \sqrt { 6 } -2 } =\frac{\sqrt{5}(\sqrt{6}-2)-\sqrt{5}(\sqrt{6}+2)}{(\sqrt{6}+2)(\sqrt{6}-2)}\)
\(\frac{ ̶​̶​̶​̶​̶​̶​̶​̶√̶3̶0̶-2\sqrt{5}- ̶​̶​̶​̶​̶​̶​̶​̶√̶3̶0̶-2\sqrt{5}}{\sqrt{6}^2-2^2}=\frac{-4\sqrt{5}}{6-4}=\frac{ ̶4̶\sqrt{5}}{ ̶2̶}=-2\sqrt{5}\)

A = {b,c,e,g,h}
B = {a,c,d,g,i}
C = {a,d,e,g,h}
B ∩ C = {a,d,g}
A - (B ∩ C) = {b,c,e,g,h} - {a,d,g} = {b,c,e,h}...(1)
A-B = {b,c,e,g,h} - {a,c,d,g,i} = {b,e,h}
A-C = {b,c,e,g,h} - {a,d,e,g,h} = {b,c}
(A - B)⋃(A - C) = {b,c,e,h}...(2)
From (1) and (2) it is verified that
A-(B ⋂ C) = (A - B) U (A - C)

Let x = \(3.1\overline { 7 } \) = 3.1777 .......  (1)
Here period of decimal is 2, multiply equation (1) by 100
10x = 31.777 .........  (2)
(2)-(1)

\( =\frac{143}{45} \)

Distance =\(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
OA = \(\sqrt { (11-1)^{ 2 }+(2-2)^{ 2 } } \)
= \(\sqrt { { 10 }^{ 2 }+{ 0 }^{ 2 } } \)
= \(\sqrt { 100 } \) = 10
OB =\(\sqrt { (11-3)^{ 2 }+(2+4)^{ 2 } } \)
= \(\sqrt { { 8 }^{ 2 }+6^{ 2 } } =\sqrt { 64+36 } \)
= \(\sqrt { 100 } \) = 10
OC = \(\sqrt { (11-5)^{ 2 }+(2+6)^{ 2 } } \)
= \(\sqrt { { 6 }^{ 2 }+8^{ 2 } } =\sqrt { 36+64 } \)
= \(\sqrt { 100 } \) = 10
OA = OB = OC = 10 units
O is the centre of the circle passing through A, B and C.

Construction :
Step 1: Draw \(\triangle\)ABC with AB = BC = CA = 8 cm
Step 2: Construct angle bisectors of any two angles (A and B) and let them meet at I. I is the incentre of \(\triangle\)ABC.
Step 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 4: With I as centre, ID as radius draw the circle. This circle touches all the sides of triangle internally.

In \(\triangle \)ABC,
AB = 4 cm, AC = 3 cm, \(\angle\)A =  90°

Construction :
Step 1: Draw \(\triangle \)ABC with AB = 4 cm, AC = 3 cm, \(\angle\)A = 90°
Step 2: Draw perpendicular bisectors of any two sides (AB and AC) to find the mid points of AB and AC.
Step 3: Draw the medians CD and BE. Let them meet at G.
Step 4: G is the centroid of the given triangle.