\(\cfrac { cos{ 63 }^{ 0 }20' }{ sin{ 26 }^{ 0 }40' } =\cfrac { cos63^{ 0 }20' }{ cos{ 63 }^{ 0 }20' } =1\)

3 tan \(\theta\) + 4 (sec \(\theta\) \(\times\) sin\(\theta\)) = 24
\(3tan\theta +4\left( \cfrac { 1 }{ cos\theta } \times sin\theta \right) =24\)
7 tan \(\theta\) = 24
\(tan\theta =\cfrac { 24 }{ 7 } \)
hypetenuse = \(\sqrt { { 24 }^{ 2 }+49 } =\sqrt { 576+49 } =\sqrt { 625 } =25\)
\(sin\theta =\cfrac { 24 }{ 25 } ;cos\theta =\cfrac { 7 }{ 25 } ;cosec\theta =\cfrac { 25 }{ 24 } ;sec\theta =\cfrac { 25 }{ 27 } ;cot\theta =\cfrac { 7 }{ 24 } \)

s = \(\frac { 41+15+25 }{ 2 } \) = 40.5
Area = \(\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 40.5\times 0.5\times 25.5\times 15.5 } \)
=\(\sqrt { \frac { 405 }{ 10 } \times \frac { 5 }{ 10 } \times \frac { 255 }{ 10 } \times \frac { 155 }{ 10 } } =\sqrt { \frac { 5\times 9\times 9\times 5\times 5\times 3\times 17\times 5\times 31 }{ 10\times 10\times 10\times 10 } } \)
=\(\frac { 1 }{ 100 } \times 5\times 9\times 5\times \sqrt { 3\times 17\times 31 } \)

\(120=\frac { 8(50)+4(y) }{ 12 } \)
1440 = 400 + 4y
4y = 1040
\(y=\frac { 1040 }{ 4 } =260\ km\)

Centre of the circle \(=\left( \frac { 10+2 }{ 2 } ,\frac { 11+3 }{ 2 } \right) =(6,7)\)

Total number of families S = 1500
n(S) = 1569
Let P be the event of selecting a family having part time maids and F be the event of selecting a family having full time maids.
(i) Both types of maids Let P \(\cap\) F be the event of selecting a family having both types of maids.
Let (P \(\cap\) F) = 259
\(\mathrm{P}(\mathrm{P} \cap \mathrm{F})=\frac{\mathrm{n}(\mathrm{P} \cap \mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{250}{1500}=\frac{1}{6}\)
(ii) Part time maids Part time maids = only part time maid + both
= 860 + 250 = 1110 '
n(P) = 1110
\(\mathrm{p}(\mathrm{P})=\frac{\mathrm{n}(\mathrm{P})}{\mathrm{n}(\mathrm{S})}=\frac{1110}{1500}=\frac{111}{150}\)
(iii) No maids Let (P \(\cup\) P)'be the event of choosing a family not having maids and (P u F) be the event of choosing a family having part time or full time or both maids.
n(P \(\cup\) F ) = only n(P) + only n(F) + n(p \(\cap\) p) = 850 + 370 + 250
n(P\(\cup\)F) = 1480
n(P\(\cup\)F)' = n(S)-n(P\(\cup\)F)
= 1500 - 1480
n(P \(\cup\) F)' = 20
\(\mathrm{P}(\mathrm{P} \cup \mathrm{F})^{\prime}=\frac{\mathrm{n}(\mathrm{P} \cup \mathrm{F})^{\prime}}{\mathrm{n}(\mathrm{S})}=\frac{20}{1500}=\frac{1}{75}\)

Mid point of AC

Mid point of BC
\(=\left( \frac { 2a+2b }{ 2 } ,\frac { 0+2c }{ 2 } \right) \)

Distance between two points \(=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \)
\(\therefore \bar { AB } =\sqrt { { (2a-0) }^{ 2 }+({ 0-0) }^{ 2 } } =\sqrt { { 4a }^{ 2 } } =2a\)

\(\frac{22}{7}\) = 3.142857142 ...... and \(\pi\) = 3.141592653289 ........
It is an irrational number.
We usually take  \(\pi\)  as  \(\frac{22}{7}\) (a rational number only But up to 2 decimal places).
But it is not exactly equal to  \(\frac{22}{7},\) it is approximate  value.

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

\(m^2+\frac{1}{m^2}-23=(m+\frac{1}{m})^2-2-23=(m+\frac{1}{m})^2-25\)
=\((m+\frac{1}{m})^2-5^2\)
\(\left( m+\frac { 1 }{ m } +5 \right) \left( m+\frac { 1 }{ m } -5 \right) \)

\(\left( \frac { 1 }{ 27 } \right) ^{ \frac { -2 }{ 3 } }=(27)^{\frac{2}{3}}=(3^3)^{\frac{2}{3}}=3^ { ̶2̶\times\frac{2}{ ̶3̶}}=3^2=9\)

Let x = 0.57777 ......... \(\rightarrow\) (1)
10x = 5.77777 ............ \(\rightarrow\) (2)
100x = 57.7777 ........... \(\rightarrow\) (3)

 \(x={52\over 90}={26\over 45}\)
\(\therefore 0.5\overline{7}={26\over 45}\)    

By using the given shapes, students has to make different shapes like birds, house, numbers, animals, rocket etc.
These pictures are called tangram pictures.

ASSESSMENT
A= {A, S, E, M, N, T}

In \(\triangle\)PQR, PQ = 5 cm, PR = 6 cm, \(\angle\)QPR = 60°

Construction:
Step 1: Draw \(\triangle\) PQR using the given measurements PQ = 9 cm, QR = 7 cm and RP = 8 cm and construct the perpendicular bisector of any two sides (PQ and QR) to find the mid-points M and N of PQ and QR respectively.
Step 2: Draw the medians PN and RM and let them meet at G. The point G is the centroid of the given \(\triangle\)PQR.

In PQR,
PQ = 5 cm,
PR = 6 cm
\(\angle \)QPR = 60°

Construction :
Step 1: Draw △ PQR with the given measurement
Step 2: Draw perpendicular bisectors of any two sides (PQ and QR) to find the midpoints of PQ and QR.
Step 3: Draw medians PD and RE. Let them meet at G.
Step 4: G is the centroid of the given △PQR.

Given PQ = PR
∴ ㄥR = 50° (opposite angles are equal)
Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any two sides (QR and PR) and let them meet at S. S is the circumcenter of Δ PQR.
Step 3: With S as centre SP = SQ = SR as radius. Draw the circumcircle.


Circum radius = 3.5 cm.
Construction of Orthocentre of a Triangle:
Orthocentre: The orthocentre is the point of concurrency of the altitudes of a triangle. Usually it is denoted by H.

Sample space S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting a number greater than 4
E = {5, 6}
\(P(E)=\frac { Number\ of\ favourable\ outcomes }{ Total\ number\ of\ outcomes } \)
\(P(E)=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =0.333...\)

Side of a cube = 7 cm
Now length of the resulting cuboid (l) = 7+7 =14 cm
Breadth (b) = 7 cm, Height (h) = 7 cm
So, Total Surface Area = 2(lb + bh + lh)
= 2[(14 \(\times\) 7)+(7 \(\times\) 7)+(14 \(\times\) 7)]

= 2(98 + 49 + 98)
= 2 × 245
= 490 cm²
Lateral Surface Area = 2(l + b) × h
= 2(14 + 7) × 7 = 2 × 21× 7
= 294 cm²

(i) \(\frac { sin\ 49° }{ cos\ 41° } \) 
sin 49⁰ = sin(90⁰ - 41⁰) = cos 41⁰,  since 49⁰ + 41⁰ = 90⁰ (complementary),
Hence on substituting sin 49^o = cos41^o we get, \( \frac { cos\ 41° }{ cos\ 41° } \)= 1
(ii) \(\frac { sec\ 63° }{ cosec\ 27° } \)
sec63^o = sec (90^o- 27^o) = cosec27^o, here, 63^o and 27^o are complementary angles
we have \(\frac { sec\ 63° }{ cosec\ 27° } =\frac { cosec\ 27° }{ cosec\ 27° } =1\)

(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 the coordinates of A be (x₁, y₁) and the given point is B(5,−6). Since the centre is the mid-point of the diameter AB, we have
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \) = 3
x₁ + 5 = 6
x₁ = 6 − 5
x₁ = 1
\(\frac { { y}_{ 1 }+{ y }_{ 2 } }{ 2 } \) = -4
y₁ − 6 = −8
y₁ = −8 + 6
y₁ = −2

Therefore, the coordinates of A is (1, −2) .

(x - 7)² = ( x - 7) (x - 7)
(x - 7)² = ( x + 7) (x + 7)
(x - 4)³ = (x - 4) (x - 4) = x(x - 3) - 4(x - 3) = (x - 3)(x - 4)
There is no common factor other than one.
Therefore, GCD = 1

The order of the surds \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 4 ]{ 7 } \) and \(\sqrt [ 2 ]{ 6 } \) 3, 4, 2 L.C.M of 3, 4, 2 = 12
\(\sqrt [ 3 ]{ 5 } ={ 5 }^{ \frac { 1 }{ 3 } }={ 5 }^{ \frac { 4 }{ 12 } }=\left( 625 \right) ^{ \frac { 1 }{ 12 } }\)
\(\sqrt [ 4 ]{ 7 } ={ 7 }^{ \frac { 1 }{ 4 } }=7^{ \frac { 3 }{ 12 } }=\left( 343 \right) ^{ \frac { 1 }{ 12 } }\)
\(\sqrt [ 2 ]{ 6 } ={ 6 }^{ \frac { 1 }{ 2 } }=7^{ \frac { 6 }{ 12 } }=\left( 46656 \right) ^{ \frac { 1 }{ 12 } }\)
The order of the surds \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 4 ]{ 7 } ,\sqrt [ 2 ]{ 6 } \)  is \(\left( 343 \right) ^{ \frac { 1 }{ 12 } }<\left( 625 \right) ^{ \frac { 1 }{ 12 } }<\left( 46656 \right) ^{ \frac { 1 }{ 12 } }\) that is \(\sqrt [ 4 ]{ 7 } <\sqrt [ 3 ]{ 5 } <\sqrt [ 2 ]{ 6 } \)

11, 12,14,18, x+12, x+4, 30, 32, 35, 41
The number of values = 10,which is an even number
Median = Average of,\(\left( \cfrac { 10 }{ 2 } \right) ^{ th }\) and \(\left( \cfrac { 10 }{ 2 } +1 \right) ^{ th }\)
\(24=\cfrac{x+2+x+4}{2}\)
\(24=\cfrac { 2x+6 }{ 2 } =\cfrac { 2\left( x+3 \right) }{ 2 } \)
\(\therefore\) x+3 = 24
x = 24-3 = 21

\(\frac { \sqrt { 7 } -2 }{ \sqrt { 7 } +2 } =a\sqrt{7}+b\)
L.H.S = \(\frac{\sqrt{7}-2\times\sqrt{7}-2}{\sqrt{7}+2\times\sqrt{7}-2}=\frac{(\sqrt{7}-2)^2}{\sqrt{7}^2-2^2}=\frac{\sqrt{7}^2-2\sqrt{7}\times2+2^2}{7-4}\)
=\(\frac{7-4\sqrt{7}+4}{3}=\frac{11-4\sqrt{7}}{3}=\frac{11}{3}-\frac{4\sqrt{7}}{3}\)
\(\frac{-4\sqrt{7}}{3}+\frac{11}{3}=a\sqrt{7}+b\)
a √7=\(\frac{-4\sqrt{7}}{3}\)
a = \(\frac{-4}{3}\)
b = \(\frac{11}{3}\)

Let M be the set of students opting for Mathematics.
Let S be the set of students opting for Science.
n (M U S) = 40, n (M) = 26, n (S) = 24
n (M U S) = n (M) + n (S) - n (M ∩ S)
40 = 26 + 24 - n (M ∩ S)
n (M ∩ S) = 26 + 24 - 40
= 50 - 40 = 10
∴ Number of students opted for Mathematics and Science = 10

Let A be the set of people owned car A
Let B be the set of people owned car B
n( A) = 400, n(B) = 200, n(A \(\cap\) B) = 50
n(A \(\cup\) B) = 500 .........(1)
n(A) + n(B) - n(A \(\cap\) B) = 400 + 200 - 50
= 600 - 50
= 550  ...........(2)
From (1) and (2) we get
\(n(A\cup B)\ne n(A) +n(B)-n(A\cap B)\)
\(\therefore\) The given data is incorrect.

Let x = 0.00010001 ............. \(\rightarrow\) (1)
10000x = 1.00010001 .......... \(\rightarrow\) (2)
(2) - (1) \(\Rightarrow\) 10000x - x = 1.00010001 ......... (-)
                                   0.00010001 .........
                                 ____________
                     9999x = 1
      \(x=\frac{1}{9999}\)      

Let the point A be (a, a), B is (1, 3)
Distance AB = 10 (Given)
By distance formula \(\sqrt { (a-1)^{ 2 }+(a-3)^{ 2 } } =10\)
Simplifying
2a²- 8a + 10 = 100
a²- 4a - 45 = 0
(a - 9)(a + 5) = 0
⇒ a = -5 ; A = (-5,-5)
a = 9 ; A = (9,9)

The given system of equations are
3x − 4y = 10 \(\Rightarrow\) 3x − 4y −10 = 0 .....(1)
  4x + 3y = 5 \(\Rightarrow\) 4x + 3y − 5 = 0 .....(2)
For the cross multiplication method, we write the coefficients as

\(\frac { x }{ (-4)(-5)-(3)(-10) } =\frac { y }{ (-10)(4)-(-5)(3) } =\frac { 1 }{ (3)(3)-(4)(-4) } \)
\(\frac { x }{ (20)-(-30) } =\frac { y }{ (-40)-(-15) } =\frac { 1 }{ (9)-(-16) } \)
\(\frac { x }{ 20+30 } =\frac { y }{ -40+15 } =\frac { 1 }{ 9+16 } \)
\(\frac { x }{ 50 } =\frac { y }{ -25 } =\frac { 1 }{ 25 } \)
Therefore, we get \(x=\frac { 50 }{ 25 } ;y=\frac { -25 }{ 25 } \)
x = 2; y = -1
Thus the solution is x = 2, y = –1.
Verification:
3x–4y = 10 ...(1)
3(2)–4(–1) = 10
6 + 4 = 10
10 = 10 True
4x + 3y = 5 ...(2)
4(2) + 3(–1) = 5
8–3 = 5
5 = 5 True

P (x) = 5x³ + 7x² + 3x + 2
d(x) = 3x + 2
Standard form of p (x) = 5x³ + 7x² + 3x + 2
and d (x) = 3x + 2

5x³ + 7x² + 3x + 2 = \(\left( x+\cfrac { 2 }{ 3 } \right) \left( 5{ x }^{ 2 }-{ \cfrac { 11 }{ 3 } x+\cfrac { 5 }{ 9 } } \right) +\cfrac { 44 }{ 27 } =\left( \cfrac { 3x+2 }{ 3 } \right) \left( 5x2-\cfrac { 11 }{ 3 } x+\cfrac { 5 }{ 9 } \right) +\cfrac { 44 }{ 27 } \)
= \(\left( \cfrac { 3x+2 }{ 3 } \right) 3\left( \cfrac { 5 }{ 3 } { x }^{ 2 }-\cfrac { 11 }{ 9 } x+\cfrac { 5 }{ 27 } \right) +\cfrac { 44 }{ 27 } \)
= \(\left( 3x+2 \right) \left( \cfrac { 5 }{ 3 } { x }^{ 2 }-\cfrac { 11 }{ 9 } x+\cfrac { 5 }{ 27 } \right) +\cfrac { 44 }{ 27 } \)
Hence the quotient \(\cfrac { 5 }{ 3 } { x }^{ 2 }-\cfrac { 11 }{ 9 } x+\cfrac { 5 }{ 21 } \) and remainder  IS \(\cfrac { 44 }{ 27 } \)