cot (90° - 75) cot (90° - 60°) cot 45° cot 60° cot 75°
= tan 75° tan 6.0° (1) cot 60° cot 75° = 1

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

(i) TSA = 6a² = 6 \(\times\) 27² = 6 \(\times\) 729 = 4374 cm²
(ii) TSA = 6a² = 6 \(\times\)3² = 54 cm²
(iii) TSA = 6a² = 6 \(\times\) 6² = 6 \(\times\) 36 = 216 cm²
(iv) TSA = 6a² = 6 \(\times\) 2.1² = 6 \(\times\) 4.41 = 26.46 cm²

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

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

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 fuli 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
\(\mathrm{P}(\mathrm{P} \cup \mathrm{F})^{\prime}=\frac{\mathrm{n}(\mathrm{P} \cup \mathrm{F})^{\prime}}{\mathrm{n}(\mathrm{S})}=\frac{20}{1500}=\frac{1}{75}\)

Two rational numbers between the given two irrational numbers are 0.5152 and 0.5352

Mid point of AC

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

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

(i)  \(\frac { 1 }{ 4 } \) and \(1\over 5\)
Take LCM = 200
\( \frac{1}{4} \times \frac{50}{50}=\frac{50}{200} \)
\( \frac{1}{5} \times \frac{40}{40}=\frac{40}{200}\)
The numbers are \(\frac{41}{200}, \frac{42}{200} \ldots \ldots \ldots \frac{49}{200}\)
(ii) 0.1 and 0.11
Change decimal into fraction \(0.1=\frac{1}{10} \text { and } 0.11=\frac{11}{100}\)
Take LCM = 100
\(\frac{1}{10} \times \frac{10}{10}=\frac{10}{100}, \frac{11}{100} \times \frac{1}{1}=\frac{11}{100}\)
Take LCM = 600
\(\frac{10}{100} \times \frac{6}{6}=\frac{60}{600} \text { and } \frac{11}{100} \times \frac{6}{6}=\frac{66}{600}\)
A rationalnumberbetween and \(\frac{60}{600} \text { and } \frac{66}{600}\)
are \(\frac{61}{600}, \frac{62}{600}, \frac{63}{600}, \frac{64}{600}, \frac{65}{600}\)
(iii) -1 and -2
Take LCM = 10 
\(\frac{-1}{1} \text { and } \frac{-2}{1}\)
\(\frac{-1}{1} \times \frac{10}{10}=\frac{-10}{10} \text { and } \frac{-2}{1} \times \frac{10}{10}=\frac{-20}{10}\)
The Numbers are \(\frac{-11}{10}, \frac{-12}{10} \ldots \ldots \ldots \ldots . \frac{-19}{10}\)

In this example, three values 20, 21, 22 occur two times each. There are three modes for the given data!

(x-2y+3z) (x²+4y²+9z²+2xy+6yz-3xz)
(a+b+c) (a²+b²+c²- ab - bc - ca) = a³+b³+c³\(\times\)3abc
ஃ (x-2y+3z) (x²+4y²+9z²+2x)
= x³- 8y³+27z³+18xyz

\({7\over 16}={7\over 2^4}={7\over 2^4\times 5^6}\)
\(\therefore {7\over 16}\) has a terminating decimal expansion.

n(X) = 10
The number of subsets of X = n[P(X)]
= 2^m
= 2¹⁰ = 1024
number of proper subsets of X = 2^m-1
= 1024 - 1
= 1023

Steps of construction:
Step 1: Draw the ΔABC with the given measure.
Step 2: Construct the perpendicular bisector of any two sides (AB and AC) and let them meet at S which is the circumcentre.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, Band C.

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

write 64⁰34' = 64⁰30' + 4'
From the table we have, sin64⁰30' = 0.9026 

Let a = 28 m, b = 15 m and c = 41 m
Then, s = \(\frac{a+b+c}{2}=\frac{28+15+41}{2}=\frac{84}{2}\) = 42m
Area of triangular field =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 42(42-28)(42-15)(42-41) } \)
=\(\sqrt { 42\times 14​​\times 17\times 1 } \)
=\(\sqrt { 2\times 3\times 7\times 7\times 2\times 3\times 3\times 3\times 1 } \)
\(=2 \times 3 \times 7 \times 3\)
= 126 m²
Given the cost of levelling is Rs. 20 per m².
The total cost of levelling the field = 20 \(\times\)126 = Rs. 2520.

Let P, Q, R be the points on the line segment joining A(−11, 4) and B(9, 8) such that AP = PQ = QR = RB .
Here Q is the mid-point of AB, P is the mid-point of AQ and R is the mid-point of QB.
Q is the mid-point of AB =\(\left( \frac { -11+9 }{ 2 } ,\frac { 4+8 }{ 2 } \right) =\left( \frac { -2 }{ 2 } ,\frac { 12 }{ 2 } \right) \)= (-1, 6)
P is the mid-point of AQ =\(\left( \frac { -11-1 }{ 2 } ,\frac { 4+6 }{ 2 } \right) =\left( \frac { -12 }{ 2 } ,\frac { 10 }{ 2 } \right) \) = (-6, 5)
R is the mid-point of QB =\(\left( \frac { -1+9 }{ 2 } ,\frac { 6+8 }{ 2 } \right) =\left( \frac { 8 }{ 2 } ,\frac { 14 }{ 2 } \right) \) = (4, 7)
Hence the points which divides AB into four equal parts are P(–6, 5), Q(–1, 6) and R(4, 7).

(7x+2y)² = (7x)² + 2(7x) (2y) + (2y)² = 49x² + 28xy + 4y²

\(9\sqrt { 7 } -6\sqrt { 7 } =\left( 9-6 \right) \sqrt { 7 } =3\sqrt { 7 } \) The answer is irrational.

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

(i) \( (300000)^{2} \times(20000)^{4}\)
\(=\left(3.0 \times 10^{5}\right)^{2} \times\left(2.0 \times 10^{4}\right)^{4} \)
\(=3^{2} \times 10^{10} \times 2^{4} \times 10^{16} \)
\(=9 \times 16 \times 10^{10+16} \)
\(=144 \times 10^{26} \)
\(=1.44 \times 10^{28} \)
(ii) (0.000001)¹¹ ÷ (0.005)³
\( =\left(1.0 \times 10^{-6}\right)^{11} \div\left(5.0 \times 10^{-3}\right)^{3} \)
\( =\frac{1.0 \times 10^{-66}}{125.0 \times 10^{9}} \)
\( =\frac{1000.0 \times 10^{-69}}{125.0 \times 10^{-9}} \)
\( =8.0 \times 10^{-69} \times 10^{9} \)
\( =8.0 \times 10^{-60} \)
(iii) \(\left\{(0.00003)^{6} \times(0.00005)^{4}\right\} \div\left\{(0.009)^{3} \times(0.05)^{2}\right\}\)
=\(\frac{(3.0\times10^{-5})^6\times(5.0\times10^{-5})^4}{(9.0\times10^{-3})^3\times(5.0\times10^{-2})^2}\)
=\(\frac{3^6\times10^{-30}\times5^4\times10^{20}}{9^3\times10^{-9}\times5^{2}\times10^{-4}}=\frac{3^6\times5^4\times10^{-30-20}}{(3^{2})^3\times10^{-9-4}\times5^{2}}=\frac{ ̶3̶^6̶\times5^4\times10^{-50}}{ ̶3̶^6̶\times5^2\times10^{-13}}\)
\( =5^{4-2} \times 10^{-50+13} \)
\(=5^{2} \times 10^{-37} \)
\(=25 \times 10^{-37} \)
\(=2.5 \times 10^{1} \times 10^{-37}\)
\(=2.5 \times 10^{-36} \)

(i) 
(ii) 
(iii) 
(iv) 

(i) A = {1,2,5,6,7}
(ii) B = {3,4,5,6}
(iii) A ∩ B = {5,6}
(iv) A U B = {1,2,3,4,5,6,7}
(v) A-B = {1,2,7}
(vi) B - A = {3,4}
(vii) (A-B) ∩ (B -A) = { }.

AD = 3 units
DA = 3 units (From graph)
Yes, Distance AD = Distance DA
The students are asked to do the activity with different points and verify the result.

Let x = -21.213777 ...... \(\rightarrow\) (1)
Here period of decimal is 4,multiply equation (1) by 10000
10000 x = -21213, 7.777 .............. (2)
(2) - (1)
9999 x = 212116.5640
\( x =\frac{-212116.5640}{9999} \)
\( x =\frac{-2121165640}{99990000} \)
\( =\frac{-190924}{9000}=\frac{-47731}{2250} \)

p(x) = x² - 9  q(x) = 6x² + 7x - 2
p(x) x q(x) = (x² - 9) (6x² + 7x - 2)
= 6x⁴ + 7x³ - 2x² - 54x² - 63x + 18
= 6x⁴ + 7x³ - 56x² - 63x + 18
The degree of the polynomial 4

Construction:
Step 1: Draw ΔABC with the given measurements AB = 8 cm, \(\angle\)A = 90^o and AC = 6 cm and construct the perpendicular bisector of any two sides (AB and AC) to find the mid points M and N of AB and BC respectively.
Step 2: Draw the medians (C and BN and let them meet at G. The point G is the centroid of the given ΔABC.

In \(\triangle \)LMN
LM = 7.5 cm,
MN = 5 cm,
LN = 8 cm

Construction :
Step 1: Draw \(\triangle \)LMN with LNM = 8 cm, MN = 5 cm, LM = 7.5 cm
Step 2: Construct perpendicular bisectors for any two sides (LN and MN) to find the mid points of LM and MN.
Step 3: Draw the medians LD, ME. Let them meet at G.
Step 4: G is the centroid of the triangle LMN.

(i) (cos0⁰ + sin 45⁰ + sin 30⁰) (sin 90⁰ - cos 45⁰ + cos 60⁰)
= \(\left[ 1+\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } \right] \left[ 1-\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } \right] \)
= \(\left[ \frac { 2\sqrt { 2 } +2+\sqrt { 2 } }{ 2\sqrt { 2 } } \right] \left[ \frac { 2\sqrt { 2 } -2+\sqrt { 2 } }{ 2\sqrt { 2 } } \right] =\left[ \frac { 3\sqrt { 2 } +2 }{ 2\sqrt { 2 } } \right] \left[ \frac { 3\sqrt { 2 } -2 }{ 2\sqrt { 2 } } \right] \)
= \(\frac { 18-4 }{ 4\left( \sqrt { 2 } \right) ^{ 2 } } =\frac { 14 }{ 4\times 2 } =\frac { 7 }{ 4 } \) 
(ii) tan²60⁰ - 2tan²45⁰ - cot²30⁰ +2sin²30⁰ + \(\frac { 3 }{ 4 } \) cosec²45⁰
= \(\left( \sqrt { 3 } \right) ^{ 2 }-2(1)^{ 2 }-\left( \sqrt { 3 } \right) ^{ 2 }+2\left( \frac { 1 }{ 2 } \right) ^{ 2 }+\frac { 3 }{ 4 } \left( \sqrt { 2 } \right) ^{ 2 }\)
= \(3-2-3+\frac { 1 }{ 2 } +\frac { 3 }{ 2 } \)
= -2 + \(\frac { 4 }{ 2 } \) = -2 + 2 = 0

Let T be the set of families buy Tamil news Papers
Let E be the set of families buy English news Papers
n(TUE) = 100
n(T) = 65 and n(E) = 55
Let "X" be the number families buy both newspapers.
By using venn-diagram
From the venn-diagrarn we get
65 - x + x + 55 - x = 100
120 - x = 100
120 -100 = x
20 = x
(i) Families buy both newspapers = 20
(ii) Families buy Tamil newspapers only = 65 - x
= 65 - 20
= 45
(iii) Families buy English newspapers only = 55 - x
= 55 - 20
= 35.

p(x) 5x³ - 9x² + 10x + 2
d (x) = x + 2
Standard form ofp (x) 5x² - 9x² + 10x + 2 and
d (x) = x + 2

5x³ - 9x² + 10x + 2 = (x + 2) (5x² - 19x + 48) - 94
Hence the quotient is 5x² - 19x + 48 and remainder is - 94