\( \frac{\tan 25^{\circ}}{\cot 65^{\circ}}+\frac{\sin 40^{\circ}}{\cos 50^{\circ}} \)
\(= \frac{\tan \left(90^{\circ}-65^{\circ}\right)}{\cot 65^{\circ}}+\frac{\sin \left(90^{\circ}-50^{\circ}\right)}{\cos 50^{\circ}} \)
\(= \frac{\cot 65^{\circ}}{\cot 65^{\circ}}+\frac{\cos 50^{\circ}}{\cos 50^{\circ}}=1+1=2 \)

\(x=\sqrt { { 13 }^{ 2 }-{ 12 }^{ 2 } } \)
\(\sqrt { 169-144 } =\sqrt { 25 } =5\)
\(sin\theta =\cfrac { 5 }{ 13 } ;cos\theta =\cfrac { 12 }{ 13 } ;tan\cfrac { 5 }{ 12 } ;cosec\theta =\cfrac { 13 }{ 5 } ;\)
\(sec\theta =\cfrac { 13 }{ 12 } ;cot\theta =\cfrac { 12 }{ 5 } \)

(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

\((x,y)=\left( \frac { 3(1)+5(-11) }{ 5+1 } ,\frac { 1(7)+5(-2) }{ 6 } \right) \)
\(=\left( \frac { 3-55 }{ 6 } ,\frac { 7-10 }{ 6 } \right) =\left( \frac { -52 }{ 6 } ,\frac { -3 }{ 6 } \right) =\left( \frac { -26 }{ 3 } ,\frac { -1 }{ 2 } \right) \)

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

When two dice are rolled
Sample space
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4),(4,5),(4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} 
n(S) = 36
(i) Event of the sum is equal to 1 = 0
∴ Probability = \(\frac { 0 }{ n(S) } \) = 0
(ii) Event of the sum is equal to 4
B = {(1, 3), (2, 2), (3, 1)}
n(B) = 3
P(B)= \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 12 } \)
(iii) Event of the sum is equal to less than 13
C= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} 
n(C) = 36
P(C) = \(\frac { n(C) }{ n(S) } =\frac { 36 }{ 6 } \).

Three irrational numbers between 0.12 and 0.13 are 0.12010010001…, 0.12040040004…, 0.12070070007…

In squares the diagonals are equal and bisect each other
∴ Mid point of BD = Mid point of AC
\(\left( \frac { -1+x }{ 2 } ,\frac { -2+y }{ 2 } \right) =\left( \frac { -5+5 }{ 2 } ,\frac { 4+2 }{ 2 } \right) \)
\(\frac { -1+x }{ 2 } =\frac { 0 }{ 2 } \quad \frac { -2+y }{ 2 } =\frac { 6 }{ 2 } \)
-1 + x = -2 + y = 6
x =1  y = 8
∴ The vertex D(x, y) = (1, 8)

(i) Given that a = 2  + \(\sqrt { 3 } \) , b = 2 - \(\sqrt { 2 } \) 
x = a + b  = ( 2 + \(\sqrt { 3 } \)) + (2 - \(\sqrt { 3 } \)) = 4 a rational number
y = a - b = (2 +  \(\sqrt { 3 } \) ) - (2 - \(\sqrt { 3 } \)) = 2\(\sqrt { 3 } \)  an irrational number
(ii) Given that a  = \(\sqrt { 2 } \) + 7 , b = \(\sqrt { 2 } \) - 7
x = a + b = (\(\sqrt { 2 } \) + 7) + (\(\sqrt { 2 } \) - 7) = 2\(\sqrt { 2 } \) an irrational number.
y = a -b  = (\(\sqrt { 2 } \) + 7 ) - (\(\sqrt { 2 } \) - 7) = 14 a rational number
(iii) Given that a = \(\sqrt { 75 } \) , b = \(\sqrt { 3 } \)
x = ab = \(\sqrt { 75 } \times \sqrt { 3 } =\sqrt { 75\times 3 } =\sqrt { 5\times 5\times 3\times 3 } =5\times 3=15\) a rational number
\(y=\frac { a }{ b } =\frac { \sqrt { 75 } }{ \sqrt { 3 } } =\sqrt { \frac { 75 }{ 3 } } =\sqrt { 25 } =5\), rational number
(iv) Given that a = \(\sqrt { 18 } \), b = \(\sqrt { 3 } \)
\(x=ab=\sqrt { 18 } \times \sqrt { 3 } =\sqrt { 18\times 3 } =\sqrt { 6\times 3\times 3 } =3\sqrt { 6 } ,\) an irrational number,
\(y=\frac { a }{ b } =\frac { \sqrt { 18 } }{ \sqrt { 3 } } =\sqrt { \frac { 18 }{ 3 } } =\sqrt { 6 } \) an irrational number

\(\sqrt { 27 } \) =\(\sqrt { 3 } \times \sqrt { 3 } \times \sqrt { 3 } =\left( { 3 }^{ \cfrac { 1 }{ 2 } } \right) ^{ 3 }=\left( 3 \right) ^{ \left( \cfrac { 3 }{ 2 } \right) }\)

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

(2a-3b+4c)²= (2a+(-3b)+4c)² = (2a)²+(-3b)²+(4c)²+2(2a)(-3b)+2(-3b)(4c)+2(4c)(2a)
= 4a²+9b²+16c²+(-12ab)+(-24bc)+16ca
= 4a²+9b²+16c²-12ab-24bc+16ca

Equivalent sets

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}
\(P(E)=\frac { Number\ of\ favourable\ outcomes }{ Total\ number\ of\ outcomes } \)
\(P(E)=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =0.333...\)

Given the dimensions of the cuboid;
that is length (l) = 7.5 m, breadth (b) = 3 m and height (h) = 5 m.
TSA = 2(lb + bh + lh)

= 2[(7.5 × 3) + (3 × 5) + (7.5 × 5)]
= 2(22.5 +15 + 37.5)
= 2 × 75
= 150 m²
LSA = 2(l + b) × h
= 2(7.5 + 3) × 5
= 2 ×10.5 × 5
= 105 m²

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

Given linear equations are
kx + 2y = 3 ......(1)
2x - 3y = 1  .......(2)
\(\left[ \begin{matrix} { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0 \\ { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0 \end{matrix} \right] \)
Here a₁ = k, b₁ = 2, a₂ = 2, b₂ = −3 ;
For unique solution we take \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \); therefore \(\frac { k }{ 2 } \neq \frac { 2 }{ -3 } \);\(k\neq \frac { 4 }{ -3 } \), that is \(k\neq -\frac { 4 }{ 3 } \)

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) .

From (1) and (2), it is verified that \(\left( A\cup B \right) '\)=\(A'\cap B'\)

\(\sqrt [ 5 ]{ 11 } +\sqrt [ 7 ]{ 11 } =\left( 5+7 \right) \sqrt { 11 } =12\sqrt { 11 } \)

47, 53, 62, 71, 83, 21, 43, 47, 41
Ascending order = 21, 41, 43, 47, 53, 62, 71, 83
The number of value 9,which is odd
\(\therefore\) Median =\(\left( \cfrac { 9+1 }{ 2 } \right) ^{ th }\quad \)
\(=\left( \cfrac { 10 }{ 2 } \right) ^{ th }\) Value = 5^th Value = 47

Let A be the required number to be added.
(x⁴ - 3x² + 2x + 6 ) + A = x⁴ - 2x³ - x + 8
\(\therefore\) A = x⁴- 2x³ - x + 8 - (x⁴ - 3x² + 2x + 6)
= x⁴ - 2x³ - x + 8 - x⁴ + 3x² - 2x - 6
= -2x³ + 3x² - 3x + 2
Hence -2x³ + 3x² - 3x + 2 must be added.

(i) A = {1,3,5,7} ; B = {1,2,4,6,8}
A-B = {1,3,:5,7} - {1,2,4,6,8}
= {3,5,7}
B -A = {1,2,4,6,S} - {1,3,5,7}
= {2,4,6,8}
AΔB = (A - B)U(B-A)
= {3,5,7} U {2,4,6,8}
= {2,3,4,5,6,7,8}.
(ii) BM = (B-A) U (A-B)
= {2,4,6,8} u {3,5,7}
= {2,3,4,516,7,S}

\(\frac { 23 }{ 10 } \)is 2.3 and\(\frac { 12 }{ 5 } \) is 2.4
You need an irrational number greater than 2.3 but less than 2.4
One such irrational number is
2.301001000100001000001000000100000001……..
observe that it is non-terminating and non-recurring.

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

Distance = \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-{ y }_{ 1 })^{ 2 } } \)
AB= \(\sqrt { (7-3)^{ 2 }+(2-2)^{ 2 } } \)
= \(\sqrt { { 4 }^{ 2 }+0^{ 2 } } \)
= \(\sqrt { 16 } =4\)
BC = \(\sqrt { (7-7)^{ 2 }+(5-2)^{ 2 } } \)
= \(\sqrt { 0+(3)^{ 2 } } \)
= \(\sqrt { 9 } =3\)
AC = \(\sqrt { (7-3)^{ 2 }+(5-2)^{ 2 } } \)
= \(\sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } \)
= \(\sqrt { 16+9 } =\sqrt { 25 } =5\)
Perimeter of ΔABC = AB + BC + AC
= 4 + 3 + 5 =12 units

In the given figure AB II CD, AD is the transversal
\(\angle CDA=\angle BAD\)
= 53° (alternate angles are equal)
In \(\triangle ECD,\angle D=\angle A=53^0\) (Alternate angles are equal)
\(\angle E+\angle C+\angle D=180^0\) (sum of the angles of a triangle)
x° + 38° + 53° = 180°
x° = 180°- 91°​​​​​​​ = 89°​​​​​​​
x = 89°

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)

(i) (x³+x²-7x-3) ÷ (x-3)
Let p (x) = x³+x²-7x-3
q (x) = x - 3
To find the zero of x - 3 :
p(x) in standard form ((i.e.) descending order)
Co-efficients are 1 1 -7 -3

Quotient is:  x² +4x+5
Remainder is :12
(ii) (x³+2x²-x-4) ÷ (x+2)
p(x) = x³+2x²-x-4
Co-efficients are 1 2 -1 -4
To find zero of x+2, put x+2 = 0; x = -2

Quotient is: (x²-1)
Remainder is: -2
(iii) (3x³-2x²+7x-5) ÷ (x+3)
To find zero of the divisor (x + 3), put x + 3 = 0;  x = - 3
Dividend in Standard form 3x³-2x²+7x-5
Co-efficients are 3 -2 7 -5
Synthetic Division

Quotient is: 3x²-11x+40
Remainder: is -125
(iv) (8x⁴-2x²+6x+5) ÷ (4x+1)
To find zero of the divisor 4x + 1, put 4x + 1 = 0; 4x = -1; x = \(-\frac{1}{4}\)
Dividend in Standard form 8x⁴-2x²+6x+5
Co-efficients are 8 0 -2 6 5
Synthetic Division

8x⁴-2x²+6x+5 = \((x+\frac{1}{4})(8x^3-2x^2-\frac{3x}{2}+\frac{51}{8})+\frac{109}{32}\)
=\(\frac{4x+1}{ ̶4̶}\times ̶4̶(2x^3-\frac{x^2}{2}-\frac{3x}{8}+\frac{51}{32})+\frac{109}{32}\)
= (4x+1)\((2x^3-\frac{x^2}{2}-\frac{3x}{8}+\frac{51}{32})+\frac{109}{32}\)
Quotient is: \(2x^2-{x^2\over2}-{3x\over8}+{51\over32}\)
Remainder is: \({109\over32}\)

Steps for construction:
Step 1: Draw the ΔABC with the given measurements.
Step 2: Construct altitudes from any two vertices A and C to their opposite sides BC and AB respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given triangle ABC.

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.