\(\frac { cos70° }{ sin20° } +\frac { cos59° }{ sin31° } +\frac { cos\theta }{ sin\left( 90°-\theta \right) } -8cos^{ 2 }60°\)
= \(\cfrac { cos\left( { 90 }^{ 0 }-{ 20 }^{ 0 } \right) }{ sin{ 20 }^{ 0 } } +\cfrac { cos\left( { 90 }^{ 0 }-{ 31 }^{ 0 } \right) }{ sin{ 31 }^{ 0 } } +\cfrac { cos\theta }{ cos\theta } -8\left( \cfrac { 1 }{ 2 } \right) ^{ 2 }\)
= \(\cfrac { { sin20 }^{ 0 } }{ { sin20 }^{ 0 } } +\cfrac { sin31^{ 0 } }{ sin{ 31 }^{ 0 } } +\cfrac { cos\theta }{ cos\theta } -8\left( \cfrac { 1 }{ 2 } \right) ^{ 2 }\)
= \(\cfrac { { sin20 }^{ 0 } }{ { sin20 }^{ 0 } } +\cfrac { sin31^{ 0 } }{ sin{ 31 }^{ 0 } } +\cfrac { cos\theta }{ cos\theta } -8\times \cfrac { 1 }{ 4 } =1+1+1-2=1\)

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

The given system of equation can be written as
2x + 7y − 5 = 0
−3x + 8y +11 = 0
For the cross multiplication method, we write the coefficients as

\(\frac { x }{ (7)(11)-(8)(-5) } =\frac { y }{ (-5)(-3)-(11)(2) } =\frac { 1 }{ (2)(8)-(-3)(7) } \)
\(\frac { x }{ 77+40 } =\frac { y }{ 15-22 } =\frac { 1 }{ 16+21 } \)
\(\frac { x }{ 117 } =\frac { y }{ -7 } =\frac { 1 }{ 37 } \)
\(\frac { x }{ 117 } =\frac { 1 }{ 37 } ,\frac { y }{ -7 } =\frac { 1 }{ 37 } \)
Hence the solution is \(\left( \frac { 117 }{ 37 } ,\frac { -7 }{ 37 } \right) \)
Verification:
2x + 7y -5 = 0 .....(1)
\(2\left( \frac { 117 }{ 37 } \right) +7\left( \frac { -7 }{ 37 } \right) -5=0\)
\(\frac { 234 }{ 57 } -\frac { 49 }{ 37 } -5=0\)
\(\frac { 185 }{ 37 } -5=0\)
5 - 5 =0 True
3x+8y+11 = 0 ...(2)
\(3\left( \frac { 117 }{ 37 } \right) +8\left( \frac { -7 }{ 37 } \right) +11=0\)
\(\frac { -351 }{ 37 } -\frac { 56 }{ 37 } +11=0\)
\(\frac { -407 }{ 37 } +11=0\)
-11 + 11 = 0 True

(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

(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).

(i) \(2\over 7\) = 0.2857142....
= \(0.\overline {285714}\)

Non-terminating and recurring decimal expansion.
(ii) \(-5\frac { 3 }{ 11 } \) = -5.272....
= \(-5.\overline { 27 } \)   

Non-terminating and recurring decimal expansion.
(iii) \(\frac { 22 }{ 3 } \) = 7.333...
= \(7.\overline { 3 } \)    

Non-terminating and recurring decimal expansion.
(vi) \(\frac{327}{200}=\frac{327}{2\times 100}\)
= \(\frac{3.27}{2}\)
= 1.635

Terminating decimal expansion.

n(A - B) = 32 + x, n(B A) = 5x
n(A \(\cap\) B) = x
From the Venn diagram:
Given n(A) = n(B)
32 + x + x = x + 5x
32 + 2x = 6x
32 = 6x - 2x
32 = 4x
x = \({32\over 4}=8\)
The value of x = 8.

Let the people who like tea be T. 
Let the people who like coffee be C 
By using formula 
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(i) n(T ∩ C) = n(T) + n(C) – n(T ∪ C) = 35 + 20 – 45 = 55 – 45 = 10 
The number of people who like both coffee and tea = 10.
(ii) The number of people who do not like tea 
n(T) = n(U) – n(T) = 45 – 35 = 10
(iii) The number of people who do not like coffee 
n(C) = n(U) – n(C) = 45 – 20 = 25.

(i) x²+10x + 24 = x²+ 6x + 4x + 24
= x(x + 6) + 4(x + 6)
= (x + 6)(x + 4)

(ii) z²+ 4z -12
= z²+ 6z - 2z -12
= z(z + 6) -2(z + 6)
= (z + 6)(z - 2)

(iii) p²- 6p -16 = p²- 8p + 2p -16
= p(p - 8)+2(p - 8)
= (p - 8)(p + 2)

(iv) t²+ 72 -17t = t²-17t + 72
= t²- 9t - 8t + 72
= t(t - 9) - 8 (t - 9)
= (t - 9)(t - 8)

(v) y² -16y- 80 = y² - 20y + 4y - 80
= y(y - 20) + 4 (y - 20)
= (y - 20) (y + 4)

(vi) a² + 10a - 600
= a² + 30a - 20a - 600
= a(a + 30)- 20(a + 30)
= (a + 30)(a - 20)
​​​​​

R.H.S = \(\frac{1}{2}(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]\)
= \(\frac{1}{2}(x+y+z)[x^2-2xy+y^2+y^2-2yz+z^2-2zx+x^2]\)
= \(\frac{1}{2}[(x+y+z)[2x^2+2y^2+2z^2-2xy-2yz-2zx^2]]\)

= x³+y³+z³-3xyz = L.H.S
Hence it is verified.

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.

(i) \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 9 ]{ 4 } ,\sqrt [ 6 ]{ 3 } \)
\(5^{\frac{1}{3}}\)
ஃ The order of the surds \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 9 ]{ 4 } ,\sqrt [ 6 ]{ 3 } \) are 3,9,6
\(4^{\frac{1}{9}}\)
\(3^{\frac{1}{6}}\) l.c.m of 3,9,6 is 18
ஃ \(\frac{1}{3}=\frac{1\times6}{3\times6}=\frac{6}{18}\)
\(\frac{1}{9}=\frac{1\times2}{9\times2}=\frac{2}{18};\frac{1}{6}=\frac{1\times3}{6\times3}=\frac{3}{18}\)
\((5^{\frac{1}{3}})=5^{\frac{6}{18}}=(15625)^{\frac{1}{18}}\)
\((4^{\frac{1}{9}})=4^{\frac{2}{18}}=(4^2)^{\frac{1}{18}}=16^{\frac{1}{18}}\)
\((3^{\frac{1}{6}})=3^{\frac{3}{18}}=(3^3)^{\frac{1}{18}}=27^{\frac{1}{18}}\)
ஃ The descending order of \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 9 ]{ 4 } ,\sqrt [ 6 ]{ 3 } \) is \((15625)^{\frac{1}{18}}>(27)^{\frac{1}{18}}>16^{\frac{1}{18}}\) i.e., \(\sqrt { \sqrt { 3 } } >\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } >\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } \)
(ii) \(\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } ,\sqrt { \sqrt { 3 } } \)
The order of the surds \(\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } ,\sqrt[2] { \sqrt [2]{ 3 } } \)  are 6, 12, 4
l.c.m of 6,12,4 is 12
\(\sqrt[2]{\sqrt[3]{5}}=5^{\frac{1}{6}}=5^{\frac{1\times2}{6\times2}}=5^{\frac{2}{12}}=(5^2)^{\frac{1}{12}}=25^{\frac{1}{12}}\)
\(\sqrt[3]{\sqrt[4]{\sqrt{7}}}=7^{\frac{1}{12}};\sqrt{\sqrt{3}}=3^{\frac{1}{4}}=3^{\frac{1\times3}{4\times3}}=3^{\frac{3}{12}}=(3^3)^{\frac{1}{12}}=27^{\frac{1}{12}}\)
ஃ The ascending order of the surds
\(\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } ,\sqrt { \sqrt { 3 } } \) is \(7^{\frac{1}{12}}<25^{\frac{1}{2}}<27^{\frac{1}{2}}\), that is \(\sqrt[3]{\sqrt[4]{7}}<\sqrt[2]{\sqrt[3]{5}}<\sqrt{\sqrt{3}}\)

Total number of subscribers = 200

From the Venn diagram,
Number of members who subscribe only one magazine = a + b + c
Number of members who subscribe exactly two magazines = x + y + z
and 125 members subscribe atleast two magazines.
That is, x + y + z + r = 125         ..... (1)
Now, n(A U B U C) = 200, n(A) = 125, n(B) = 100, n(C) = 150, n(A∩B) = x + r
n(B∩C) = y + r, n(A∩C) = z + r, n(A∩B∩C) = r
We know that,
n(AUBUC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)
200 = 125 + 100 + 150 – x – r – y – r – z – r + r
= 375 - (x + y + z + r) - r
= 375 -125    [∵ x + y + z + r = 125]
200 = 250 - r    ⇒ r = 50
From (1) x + y + z + 50 = 125
We get, x + y + z = 75
Therefore, number of members who subscribe exactly two magazines = 75.
From Venn diagram, (a + b + c) + (x +y + z + r) = 200 ........ (2)
substitute (1) in (2)
a + b + c + 125 = 200
a + b + c = 75
Therefore, number of members who subscribe only one magazine = 75.

a = 600 - (120 -x + x + 80 -x)
= 600 - (200 -x)
= 600 - 200 + x = 400 + x
b = 350 - (120 -x + x + 100 -x)
= 350 - (220 - x)
= 350 - 230 + x = 130 + x
c = 280 - (80 - x + x + 100 - x)
2800 - (180 -x)
= 280 - 180 + x = 100 + x
Each farmer grew atleast one of the above three, the number of farmers who grew all the three is x.
= a + b + c + 120 - x + 100 - x + 80 - x + x = 1000
400 + x + 130 + x + 100 + x + 120 - x + 100 - x + 80 - x + x = 1000
ஃ 930 + x = 1000
x = 1000 - 930 = 70
ஃ 70 farmers grew all the three crops.

A = { y : y = \(\frac { a+1 }{ 2 } \), a\(\in \) W and a ≤ 5}
a = {0,1,2,3,4,5}
y = \(\frac{a+1}{2}=\frac{a+1}{2}=\frac{1}{2}\)
y = \(\frac{1+1}{2}=\frac{2}{2}=1\)
y = \(\frac{2+1}{2}=\frac{3}{2};y=\frac{3+1}{2}=\frac{4}{2}=2\)
y = \(\frac{4+1}{2}=\frac{5}{2};y=\frac{5+1}{2}=\frac{6}{2}=3\)
ஃ A = \(\{\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},3\}\)
B = { y : y = \(\frac{2n-1}{2}\), n∈W and n < 5}
n = {0,1,2,3,4}
y = \(\frac{2\times0-1}{2}=\frac{-1}{2}\)
y = \(\frac{2\times1-1}{2}=\frac{1}{2}; y=\frac{2\times2-1}{2}=\frac{3}{2}\)
y = \(\frac{2\times3-1}{2}=\frac{5}{2}; y=\frac{2\times4-1}{2}=\frac{7}{2}\)
ஃ B = \(\{-\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2}\}\)
C = \(\{-1,-\frac{1}{2},1,\frac{3}{2},2\}\)
BUC = \(\{-1,-\frac{1}{2},\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},\frac{7}{2}\}\)
A-(BUC) = {3}...............(1)
A-B = {1,2,3}
A-C = \(\{\frac{1}{2},\frac{5}{2},3\}\)
(A-B)∩(A-C) = {3}...................(2)
From (1) and (2), it is verified that A-(BUC) = (A-B)∩(A-C)

p(x) = ax³ - 3x² + 4
When it is divided by x - 2
p(2) = a(2)³ - 3(2)² + 4
p = 8a-12 + 4
p = 8a- 8
q(x) = 2x³ - 5x + a
When it is divided by x - 2
q(2) = 2(2)³ - 5(2) + 8
q = 16-10+a
q = 6+a
Given p - 2q = 4
8a - 8 - 2 (6 + a) = 4
8a- 8 -12 -2a = 4
6a-20 = 4
6a = 24
a = 24/6 = 4
The value of a = 4

Let f(x) = 2x⁴- 6x³+3x²+3x-2
g(x) = x² –3x + 2
= x²-2x-x+2
= x(x-2)-1(x-2)
= (x-2)(x-1)
we show that f(x) is exactly divisible by (x–1) and (x–2) using remainder theorem
f(1) = 2(1)⁴-6(1)³+3(1)²+3(1)-2
= 2 – 6 + 3 + 3 – 2 = 0
f(2) = 2(2)⁴-6(2)³+3(2)²+3(2)-2
= 32 – 48 + 12 + 6 – 2 = 0
f(x) is exactly divisible by (x – 1) (x – 2)
i.e., f(x) is exactly divisible by x² –3x + 2
If p(x) is divided by (x -a) with the remainder p(a) = 0 , then (x -a) is a factor of p(x). Remainder Theorem leads to Factor Theorem.

Distance = \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
AB = \(\sqrt { (2-1)^{ 2 }+(1-1)^{ 2 } } \)
= \(\sqrt { (1)^{ 2 }+(0)^{ 2 } } =\sqrt { 1 } =1\)
BC = \(\sqrt { (2-1)^{ 2 }+(2-1)^{ 2 } } \)
= \(\sqrt { (0)^{ 2 }+(1)^{ 2 } } =\sqrt { 1 } =1\)
CD =\(\sqrt { (1-2)^{ 2 }+(2-2)^{ 2 } } \)
= \(\sqrt { (-1)^{ 2 }+(0)^{ 2 } } =\sqrt { 1 } =1\)
AD =\(\sqrt { (1-1)^{ 2 }+(2-1)^{ 2 } } \)
= \(\sqrt { (0)^{ 2 }+(1)^{ 2 } } =\sqrt { 1 } =1\)
AB = BC = CD = AD =1
All the four sides are equal.
ABCD is a rhombus

\(1\over 11\) = 0.0909 ......... = \(0.\overline{09}\)
\(\therefore \frac{1}{33}=\frac{1}{3}\times \frac{1}{11}\)
 \(=\frac{1}{3}\times 0.0909.......\)
= 0.0303........ = \(0.\overline{03}\)
Also \(\frac{71}{33}=2\frac{5}{33}=2+\frac{5}{33}\)
\(=2+5\times \frac{1}{33}\)
= 2 + 5 \(\times\) \(0.\overline {03}\)
= 2 + (5 \(\times\) 0.030303 .......)
= 2 + 0.151515 .......
= 2 + \(0.\overline {15}\)
= \(2.\overline {15}\)

Given ∠BAD = 120° and AC bisects ∠BAD
\(\angle{BAC}=\frac{1}{2}\times 120^{0}=60^{0}\)
∠1 = ∠2 = 60°
AD || BC and AC is the traversal
∠2 = ∠4 = 60°
Δ ABC is isosceles triangle    [∴ ∠1 = ∠4 = 60 °]
⇒ AB = BC
Parallelogram ABCD is a rhombus.

Given points are A(2, 2) and B(8, –4) and let P = (x, 0) [P lies on x axis]
By the distance formula
\(d=\sqrt{(x_2 -x_1)^2 +(y_2 -y_1)^2}\)
\(AP=\sqrt{(x-2)^2+(0-2)^2}=\sqrt{X^2-4x+4+4}=\sqrt{x^2 -4x+8}\)
\(BAP=\sqrt{(x-8)^2+(0+4)^2}=\sqrt{X^2-16x+64+16}=\sqrt{x^2 -16x+80}\)
Given, AP : PB = 1 : 2
i.e \(\frac{AP}{BP}=\frac{1}{2} (\therefore BP=PB)\)
2AP = BP
squaring on both sides
4AP² = BP^{2
\(4(x^2 -16x+8)=(x^2-16x+80)\)}
\(4x^2 -16x+32=x^2-16x+80\)
3x²- 48 = 0
3x²= 48
x² = 16
x =\(\pm\)4
As the point P lies on x-axis (positive side), its x- coordinate cannot be –4.
Hence the coordinates of P is(4, 0)

To multiply (4x – 5) and (2x² + 3x – 6) distribute each term of the first polynomial to every term of the second polynomial. In this case, we need to distribute the terms 4x and –5. Then gather the like terms and combine them:

= 8x³+12x²-24x+10x²-15x+30
= 8x³ +2x²–39x +30
Aliter: You may also use the method of detached coefficients:

∴  (4x –5) (2x²+3x–6) = 8x³ + 2x² – 39x + 30

In \(\triangle ABC\angle B=60^0\ and\ \angle C=70^0\)
\(\therefore\angle A=180^0-(60^0+70^0)\)
\(=180^0-130^0\) = 50⁰
In \(\triangle E=70^0\ and \angle D=60^0\)
\(\therefore \angle F=180^0-(70^0+60^0)\)
= 180° - 130° = 50°
\(\angle A=\angle F=50^0\)
\(\angle B=\angle D=60^0\)
\(\angle C=\angle E=70^0\)
By AAA congruency
\(\therefore\triangle ABC\cong\triangle FDE\)

Step 1: 
Draw the ΔABC with the given measurements.

Step 2:
Construct the perpendicular bisector of any two sides (AC and BC) and let them meet at S which is the circumcentre.
Step 3:
S as centre and SA = SB = SC as radius, draw the Circumcircle to passes through A, B and C. Circumradius = 3.9 cm.