(i) sin \(\theta\) = 0.9858  = 0.9857 + 0.0001
From the sine table 0.9857 = 80^o 18'
Mean difference 1 = 2′
0.9858 = sin 80⁰20'
sin\(\theta\) = 0.9858 = sin 80⁰20'
\(\theta\) = 80⁰20'
(ii) cos \(\theta\) = 0.7656 =  0.7660 - 0.0004
From the natural cosine table
0.7660 = 40°0′
Mean difference 4 = 2′       
0.7656 = 40°0′         
cos \(\theta\) = 0.7656 = cos 40°2'
\(\theta\) = 40°2'                                                                                          

Let ABCD be the rhombus.
Its perimeter = 4 × side = 400 m
Therefore, each side of the rhombus = 100 m
Given the length of the diagonal AC = 120 m
In \(\triangle\)ABC, let a =100 m, b =100 m, c = 120 m

s = \(\frac{a+b+c}{2}=\frac{100+100+120}{2}\) = 160 m
Area of \(\triangle\)ABC =\(\sqrt{160(160-100)(160-100)(160-120)}\)
= \(\sqrt{160 \times 60\times 60 \times40}\)
= \(\sqrt{40 \times 2 \times \times2\times60\times60\times40}\)
= 40 × 2 × 60 = 4800 m²
Therefore, Area of the field ABCD = 2 × Area of \(\triangle\)ABC = 2 × 4800 = 9600 m²

In the quadrilateral ABCD, join one of the diagonals, say AC.
Area of \(\triangle\)ABC = \(\frac{1}{2}\)\(\times\) base \(\times\) height
=\(\frac{1}{2}\)\(\times\)8\(\times\)15\(\times\) 60 cm²
By Pythagoras theorem, in right angled triangle ABC,
AC² = AB² + BC²
= 8² +15² = 64 + 225 = 289 cm
Therefore, AC =\(\sqrt{289}\) =17cm
Now, for\(\triangle\)ACD, let us consider a = 17 cm, b =12 cm, c =25 cm
then, s = \(\frac{a+b+c}{2}=\frac{17+12+25}{2}=\frac{54}{2}\) = 27cm

Area of \(\triangle\)ACD =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt{27(27-17)(27-12)(27-25)}\)
=\(\sqrt{27\times10\times15\times2}\)
=\(\sqrt{3\times3\times3\times2\times5\times5\times3\times2}\)
= 3 × 3 × 2 × 5 = 90cm²
Therefore, Area of quadrilateral ABCD
=Area of \(\triangle\)ABC + Area of \(\triangle\)ACD
= 60 + 90 = 150 cm²

(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 the vertices of the ABC be A(x₁, y₁), B(x₂, y₂ ) and C(x₃, y₃) and the given mid-points of the sides AB, BC and CA are (5, 1), (3, −5) and (−5, −1) respectively. Therefore
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =5 \Rightarrow\)x₁ + x₂ = 10 ...(1)
\(\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =3 \Rightarrow\)x₂ + x₃ = 6 ...(2)
\(\frac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } =-5 \Rightarrow\)x₃ + x₁ = –10 ...(3)
Adding (1), (2) and (3)
2x₁ + 2x₂ + 2x₃ = 6
x₁ + x₂ + x₃ = 3 ...(4)
(4) − (2) \(\Rightarrow\) x₁ = 3 − 6 = −3 
(4) − (3) \(\Rightarrow\) x₂ = 3 +10 = 13
(4) − (1) \(\Rightarrow\) x₃ = 3 −10 = −7
\(\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =1\)\(\Rightarrow\)y₁ + y₂ = 2 …(5)
\(\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } =-5\)\(\Rightarrow\)y₂ + y₃ = –10 …(6)
\(\frac { { y }_{ 3 }+{ y }_{ 1 } }{ 2 } =1\)\(\Rightarrow\)y₃ + y₁ = –2 …(7)
Adding (5), (6) and (7),
2y₁ + 2y₂ + 2y₃ = −10
y₁ + y₂ + y₃ = −5 ...(8)
(8) − (6) \(\Rightarrow\) y₁ = −5 +10 = 5
(8) − (7) \(\Rightarrow\) y₂ = −5 + 2 = −3
(8) − (5) \(\Rightarrow\) y₃ = −5 − 2 = −7
Therefore the vertices of the triangles are A(−3, 5), B(13, −3) and C(−7, −7).

(i) Let us prepare a table to find the ordered pairs of points for the line y = 3x −1.
We shall assume any value for x, for our convenience let us take −1, 0 and 1.
When x = −1, y = 3(–1)–1 = –4
When x = 0 , y = 3(0)–1 = –1
When x = 1, y = 3(1)–1 = 2

The points (x,y) to be plotted :
(−1, −4), (0, −1) and (1, 2).

(ii) Let us prepare a table to find the ordered pairs of points for the line y =\(\left( \frac { 2 }{ 3 } \right) x+3\)
Let us assume −3, 0, 3 as x values.
(why?)
When x = -3, \(y=\frac { 2 }{ 3 } (-3)+3=1\)
When x = 0, \(y=\frac { 2 }{ 3 } (0)+3=3\)
When x =3, \(y=\frac { 2 }{ 3 } (3)+3=5\)

The points (x, y) to be plotted: (-3, 1), (0, 3) and (3, 5).

Let M and S represent the student failed in Mathematics and Science.
Given: Number of students passed in Mathematics is 50%
∴ Number of students failed in Mathematics = 100 - 50% = 50%
n(M) = 50%

Number of students passed in Science is 70%
∴ Number of students failed in Science = 100 - 70% - 30%
n(S) = 30%
Number of students failed in both the subjects is 10%
n(M\(\cap\)S) = 10%
n(M\(\cup\)S) = n(M) + n(S) - n(M\(\cap\)S)
= 50 + 30 - 10
= 80 - 10 = 70
Given: 70% of the students failed in atleast anyone of the subject
∴ 30% of the students passed in atleast anyone of the subjects.
30 students passed mean, the total number of students is 100.
∴ 300 students passed means, the total number of students = \(\frac{100 \times 30}{30}\)= 1000
Total number of students appeared in the examination is = 1000

\(\left( A\cup B\cup C \right) \) = {1,2,3,4,5,6,8}
\(\therefore \ n\left( A\cup B\cup C \right) \) =7
Also, n(A) = 5, n(B) = (C) = 4,
Further,\(A\cap B\) = {3,5,6} \(\Rightarrow \) \(n\left( A\cap B \right) \) = 3
\(B\cap C\) = {3,6} \(\Rightarrow \)  \(n\left( B\cap C \right) \) = 2
\(A\cap C\) = {3,5,6} \(\Rightarrow \) \(n\left( A\cap C \right) \) = 3
Also, \(A\cap B\cap C\) = {3,6} \(\Rightarrow \)n\(\left( A\cap B\cap C \right) \) = 2
now, \(n\left( A\cup B\cup C \right) \)= n(A) + n(B) + n(C) -\(n\left( A\cap B \right) -n\left( B\cap C \right) -n\left( A\cap C \right) -n\left( A\cap C \right) +\left( A\cap B\cap C \right) \)
7 = 5 + 4 + 4 - 3 - 2 - 3 + 2
7 = 13 - 8 + 2
7 = 7

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(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

Modal class is 16 -20 since it has the maximum frequency.
l = 15.5, f = 32, f₁ = 16, f₂ = 24, c = 20.5–15.5 = 5
Mode \(=l+{\left(f-f_1\over 2f-f_1-f_2\right)}\times c\)
\(=15.5+\left(32-16\over64-16-24\right)\times 5\)
\(=15.5+\left(16\over24\right)\times = 15.5 + 3.33 =18.83\)

Let us arrange the marks in ascending order and prepare the following data:

Here N = 41
Median = size of \(\left(N+1\over 2\right)^{th}\) value = size of \(\left(41+1\over2\right)^{th}\) value = size of 21^st value.
If the 41 students were arranged in order (of height), the 21^st student would be the middle most one, since there are 20 students on either side of him/her. We therefore need to find the height against the 21^st student. 15 students (see cumulative frequency) have height less than or equal to 154 cm. 22 students have height less than or equal to 155 cm. This means that the 21^st student has a height 155 cm.
Therefore, Median = 155 cm

Let P(x) = ax²+5x+b
(x-2) is a factor of P(x), if P(2) = 0
P(2) = a(2)²+5(2)+b = 0
4a +10+b = 0
4a+b = -10...(1)
\((x-\frac{1}{2})\) is a factor of P(x), if P\((\frac{1}{2})\) = 0
P\((\frac{1}{2})\) = a\((\frac{1}{2})^2\) + 5\((\frac{1}{2})\)+ b = 0
\(\frac{a}{4}+\frac{5}{2}+b=0\)
\(\frac{a}{4}+b=\frac{-5}{2}\)
\(\frac{a+4b}{4}=\frac{-5}{2}\)
2a + 8b = -20
a + 4b = -10...(2)
From (1) and (2)
4a + b = -10...(1)
a + 4b = -10...(2)
(1) and (2) ⇒ 4a + b = a + 4b
3a = 3b
ஃ a = b
Hence it is proved.

To represent \(-\frac { 2 }{ 11 } ,-\frac { 5 }{ 11 } and-\frac { 9 }{ 11 } \)on the number line we make 11 markings each being equal distance \(\frac { 1 }{ 11 } \) on the left of 0.
The point A represents \(\left( -\frac { 2 }{ 11 } \right) \) , the point B represents\(\left( -\frac { 2 }{ 11 } \right) \) and the point C represents \(\left( -\frac { 9 }{ 11 } \right) \)

Relational between \(\frac { 1 }{ 2 } \) and \(\frac { 1 }{ 5 } \) = \(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
=\(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
= \(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
Rational numbers between\(\frac { 1 }{ 2 } \) and \(\frac { 7 }{ 20 } \)
=\(\frac { 1 }{ 2 } \left( \frac { 10+7 }{ 20 } \right) \)
=\(\frac { 17 }{ 40 } \)
Rational number between \(\frac { 1 }{ 2 } \) and \(\frac { 7 }{ 20 } \) =\(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } \times \frac { 17 }{ 40 } \right) \)
=\(\frac { 1 }{ 2 } \left( \frac { 20+17 }{ 40 } \right) \)
=\(\frac { 37 }{ 80 } \)
Thus the rational numbers are \(\frac { 7 }{ 20 } ,\frac { 17 }{ 40 } and\frac { 37 }{ 80 } \)

\(\sqrt { 2 } \) =1.414213...
\(\sqrt { 3} \) =1.73205....
\(\frac { 7 }{ 9 } \) =0..7777
\(\frac { 6 }{ 7 } \) =0.857142
\(\frac { 23 }{ 6 } \) =3.8333...
\(\frac { 48 }{ 5 } \) =9.6
(i) Arrange the ten real numbers in the given boxes in ascending order.

(ii) Arrange the same numbers in the boxes given below in descending order.

Distance = \(\sqrt{(x_2-x_2)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(6-3)^2+(5-7)^2}\)
\(\sqrt{3^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}\)
BC=\(\sqrt{(15-6)^2+(-1-5)^2}=\sqrt{(9)^2+(-6)^2}\)
\(\sqrt{81+36}=\sqrt{117}\)
\(\sqrt{9\times 13}-3\sqrt{13}\)
AC =\(\sqrt{(15-3)^2+(-1-7)^2}\)
\(\sqrt{12^2+(-8)^2}=\sqrt{144+64}\)
\(=\sqrt{208}=\sqrt{16\times 13}=4\sqrt{13}\)

AB + BC =AC\(\Rightarrow \sqrt{13}+3\sqrt{13}=4\sqrt{13}\) 
\(\therefore\)The points A,B,C are collinear.

Let A(1, 1),B(5, 4) and C(-2, 5)
Distance =\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(5-1)^2+(4-1)^2}\)
\(=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\)
BC =\(\sqrt{(-2-5)^2+(5-4)^2}\)
\(=\sqrt{(-7)^2+1^2}=\sqrt{49+1}=\sqrt{50}\)
AC =\(\sqrt{(-2-1)^2+(5-1)^2}\)
\(=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\)

AB = 5, AC = 5,
\(\therefore\) ABC is an isosceles triangle .................(1)
BC² = AB² + AC²
50 = 25 + 25\(\Rightarrow\) 50 = 50
\(\therefore\) \(\angle\)A = 90° ...................(2)
From (1) and (2) we get ABC is an isosceles right angle triangle.

Given P (6, –1), Q (1, 3) and R (a, 8)
\(PQ=\sqrt { \left( 1-6 \right) ^{ 2 }+\left( 3+1 \right) ^{ 2 } } =\sqrt { \left( -5 \right) ^{ 2 }+\left( 4 \right) ^{ 2 } } = \sqrt { 41 } \)
\(QR=\sqrt { \left( a-1 \right) ^{ 2 }+\left( 8-3 \right) ^{ 2 } } =\sqrt { \left( a-1 \right) ^{ 2 }+\left( 5 \right) ^{ 2 } } \)
Given PQ = QR
Therefore \(\sqrt { 41 } =\sqrt { \left( a-1 \right) ^{ 2 }+\left( 5 \right) ^{ 2 } } \)
41= (a –1)² + 25 [Squaring both sides]
(a–1)² + 25 = 41
(a–1)² = 41 - 25
(a–1)² = 16
(a–1) = \(\pm \) 4 [taking square root on both sides]
a = 1  \(\pm \)4
a = 1 + 4 or a = 1 – 4
a = 5, –3

Represent \(\sqrt{6.5}\) on a number line.
Steps of construction:
1. Draw a line and mark a point A and B such that AB = 6.5 cm.
2. Mark a point C on this line such that BC = 1 cm.
3. Find the mid point of AC by drawing perpendicular bisector of AC and let it be "O".
4. With O as centre and OC = OA as radius draw a semicircle.
5. Draw a line BD, which is perpendicular to AB at B.
6. Now BD = \(\sqrt{6.5}\) which can be marked in the number line as the value of BE = BD = \(\sqrt{6.5}\).

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 \(\triangle\)ABC, AB = BC = 8 cm, \(\angle\)B = 70°.

Construction :
Step 1: Draw \(\triangle\)ABC with BC = 8 cm, \(\angle\)B = 70°. AB = 8.
Step 2: Construct the angle bisectors of any two angles (B and C) and let them meet at I. Then I is the incentre of \(\triangle\)ADC. Draw perpendicular from I to anyone of the side (BC) to meet BC at D.
Step 3: With I as centre and ID as radius draw a circle. This circle touches all the sides of the triangle internally.

Steps for construction:
Step 1: Draw the rough diagram and mark the measurements
Step 2: Draw the ΔABC with the given measurements.
Step 3: Construct altitudes from any two vertices Band C to their opposite sides AC and BC respectively.
Step 4: The point of intersection of the altitude H is the orthocentre of the given ΔABC.

Steps for construction:
Step 1: Draw the ΔPQR with the given measures.
Step 2: Construct the perpendicular bisector of (PQ and PR) any two sides and let them meet at S which is the circumcenter.
Step 3: With S as centre and SP = SQ = SR as radius draw the circumcircle to passes through P, Q and R.


Circum radius = 4.3 cm.