Let BC be the height of the tower and CD be the height of the pole
Let ‘A’ be the point of observation.
Let BC = x and AB = y.
From the diagram,
ㄥBAD = 60° and ㄥXCA = 45° = ㄥBAC
In right triangle ABC, tan 45^o = \(\frac{BC}{AB}\)
gives 1 = \(\frac{x}{y}\) so, x = y         ...(1)
In right triangle ABD, tan60° = \(\frac{BC}{AB}\) = \(\frac{BC+CD}{AB}\)
gives \(\sqrt3\)  = \(\frac{x+5}{y}\) so, \(\sqrt3\)y = x + 5
we get \(\sqrt3\) x = x + 5 [From (1)]
so, \(\frac { 5 }{ \sqrt { 3 } -1 } =\frac { 5 }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } =\frac { 5(1.732+1) }{ 2 } \) = 6.83
Hence, height of the tower is 6.83 m.

\(\Delta ACD\sim \Delta ABC\)
So,
\(\cfrac { AC }{ AB } =\cfrac { AD }{ AC } \)
AC²  = AB ·AD
Similarly \(\Delta BCD\sim \Delta BAC\)
So,
\(\cfrac { BC }{ BA } =\cfrac { BD }{ BC } \)
BC²  = BA·BD
From (1) and (2)
\(\cfrac { { BC }^{ 2 } }{ AC^{ 2 } } =\cfrac { BA.BD }{ AB.AD } =\cfrac { BD }{ AD } \)

Radius of the cylinder be r
Height of the cylinder be h
Total surface area of the article
= CSA of cylinder + CSA of 2 hemispheres
= 2ㅠrh + 2πr² = 2πr(h + 2r)
\(=2\times \frac { 22 }{ 7 } \times 3.5\times (10+2\times 3.5)\)
= 22 x 17 = 374 cm²

CV =\(\frac { \sigma }{ \bar { x } } \) x 100 ⇒ \(\bar { x } =\frac { \sigma }{ CV } \) x 100
\(\bar { x } =\frac { 15.6 }{ 6.9 } \) x 100  = 22.6

1,1,1,2,2,2,3,3,3
t₂ - t₁ = 1-1 = 0
t₃ - t₂ = 1-1 = 0
t₄ - t₃ = 2-1 = 1
Here t₂ - t₁ ≠ t₃ - t₂
ஃ It is not an A.P.

f(3) =. 2x - 1
= 2(3) - 1 = 6 - 1 = 5

We know that diagonals at a parallelogram bisect each other.
So, the coordinates of the mid-point at AC = coordinates of the mid-point at BD.
i.e., \(\left[ \frac { 6+9 }{ 2 } ,\frac { 1+4 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 2+3 }{ 2 } \right] \)
\(\left[ \frac { 15 }{ 2 } ,\frac { 5 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 5 }{ 2 } \right] \)
\(\frac { 15 }{ 2 } =\frac { 8+P }{ 2 } \)
P = 7

If the battery power is 0 then y = 0
Therefore, 0 = − 0.25x + 1 gives - 0.25 x = 1 hence x = 4 hours.
Thus, after 4 hours, the battery of the mobile phone will have no power.

ln tossing a coin 3 times, the sample space
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 3
(i) Let 'A' be the event of getting double entry fee. She received double entry fee when she throws 3 heads.
A = {HHH}
n(A) = 1
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{1}{8}\)
(ii) Let 'B' be the event of getting the entry fee back. She receives her entry fee back when she throws one or two heads
B = {HHT HTH, HTT, THH, THT, TTH}
n(B) = 6
\(\mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=\frac{6}{8}=\frac{3}{4}\)
Probability of getting back entry fee =  \(\frac{3}{4}\)
(iii) Let 'C' be the event of losing the entry fee
c = {TTT}
n(C) = 1
\(\mathrm{P}(C)=\frac{n(C)}{n(S)}=\frac{1}{8}\)
Probabilify of losing the entry fee = \(\frac{1}{8}\)

\(\frac { 4{ x }^{ 2 }y }{ 2{ x }^{ 2 } } \times \frac { 6x{ z }^{ 3 } }{ 20{ y }^{ 4 } } =\frac { { 3x }^{ 3 }z }{ 5{ y }^{ 3 } } \)

Let h and r be the height and radius of the cone and cylinder.
Let l be the slant height of the cone.
Given that, h = 2.4 cm and d = 1.4 cm ; r = 0.7 cm
Here, total surface area of the remaining solid} C.S.A. of the cylinder + C.S.A. of the cone + area of the bottom
= 2\(\pi\)rh + \(\pi\)rl + \(\pi\)r² sq.units
Now, \(\\ \\ l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { 0.49+5.76 } =\sqrt { 6.25 } =2.5cm\)
Area of the remaining solid = 2\(\pi\)rh + \(\pi\)rl + \(\pi\)r² sq.units
= \(\pi\)r(2h + l + r)
\(\frac { 22 }{ 7 } \times 0.7\times [(2\times 2.4)+2.5+0.7]\)
Therefore, total surface area of the remaining solid is 17.6 m²

Given: A is the set of all natural numbers less than 8 A = {1,2,3,4,5,6,7}
B is the set of all prime numbers less than 8 B = {2,3,5,7}
'C' is the set of all even prime number
C = {2}
verify
(A ∩ B) x C = (A x C) ∩ (B x C)
\(A\cap B\) = {2,3,5,7}
\((A \cap B) \times C=\{2,3,5,7\} \times\{2\}\)
= {(2,2),(3,2),(5,2),(7,2}}    ......(1)
A X C = {1,2,3,4,5,6,7} x {2}
= {(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)}
B x C = {2,3,5,7} x {2}
= {(2,2),(3,2),(5,2),(7,2)}
\((A\times C)\cap (B\times C)\) = {(2,2) ,(3,2),(5,2),(7,2)}  ...(2)
From (1) and (2) it is clear that
(A ∩ B) x C = (A x C) ∩ (B x C)
Hence verified.

Volume of the cylindrical tank = Volume of the rectangle tank
πr²h = 28 x 16 x 11 m³

\(h=\frac { 16\times 11 }{ 88 } =2m\)

Given XY IIA C

So,

\(\therefore \Delta ABC\sim \Delta XbY\) (AAA similarity criterion)
So, \(\cfrac { ar(ABC) }{ ar(XBY) } =\left( \cfrac { AB }{ XB } \right) ^{ 2 }\)         ...(1)
ar(ABC) = 2ar(XBY)
\(\cfrac { ar(ABC) }{ ar(ABC) } =\cfrac { 2 }{ 1 } \)           ...(2)
From (1) and (2),
\(\left( \cfrac { AB }{ XB } \right) ^{ 2 }=\cfrac { 2 }{ 1 } i.e.,\cfrac { AB }{ XB } =\cfrac { \sqrt { 2 } }{ 1 } \)
\(\cfrac { XB }{ AB } =\cfrac { 1 }{ \sqrt { 2 } } \)
\(1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\cfrac { AB-XB }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)
\(\cfrac { AX }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } =\cfrac { 2-\sqrt { 2 } }{ 2 } \)

Let assumed mean A = 30
C = 5

x	d' = \(\frac { x-30 }{ 5 } \)	d'2
20	-2	4
25	-1	1
30	0	0
35	1	1
40	2	4
	Σd' = 0	Σd'2 = 10

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
To convert the values for 100, all the values will be multiplied by 2. Therefore the new values are 40, 50, 60, 70, 80.
Let A = 60, C = 10

x	d' = \(\frac { x-60 }{ 10 } \)	d'2
40	-2	4
50	-1	1
60	0	0
70	1	1
80	2	4

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 10 
=\(\sqrt { 2 } \) x 10
= 10\(\sqrt { 2 } \)
S.D. also be multiplied by 2. It is also true for the division also.

D= b²-4ac
⇒ 4(ac + bd)² - 4(a² + b²)(c² + d²)
⇒ 4[(ac + bd)² - (a² + b²)(c² + d²)]
⇒ 4(a²c² + b²d² + 2acbd - a²c²b²c² - a²d² - b²d²]
⇒ 4[2acbd - a²d² - b²c²]
⇒ 4[a²d² + b²c² - 2adbc]
⇒-4[ ad - bc]²
We have ad≠ bc
∴ ad- be of 0
⇒ (ad - bc)² > 0
⇒ 4(ad - bc)² < 0 ⇒ D < 0
Hence the given equation has no real roots.

Let us start with taking a, where a is a +ve odd integer.
We apply the division algorithm with 'a' and 'b' = 4.
Since 0  ≤ r < 4, the possible remainders are 0,1,2,3.
That is, a can be 4q, or 4q + 1, or 4q + 2 or 4q + 3, where 1 is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Any odd integer is of the form 4q + 1 or 4q + 3

Let A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) be the given paints. One way at showing that ABCD is a square is to use the property that all its sides should be equal and both its diagonals should be equal.
Now,
AB = \(\sqrt { (1-4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
BC = \(\\ \sqrt { (4+1)^{ 2 }+(2+1)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
CD = \(\sqrt { (-1+4)^{ 2 }+(-1-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
DA =\(\sqrt { (1+4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
AC = \(\sqrt { (1+1)^{ 2 }+(7+1)^{ 2 } } =\sqrt { 4+64 } =\sqrt { 68 } \)
BD = \(\\ \sqrt { (4+4)^{ 2 }+(2-4)^{ 2 } } =\sqrt { 64+4 } =\sqrt { 68 } \)
Since, AB = BC = CD = DA and AC = BD, all the four sides at the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Therefore, ABCD is a square.

x² + 7x + 10 here, a = -1, b = 7, c =10
if α and β are roots of the equation then,
α + β = \(\frac {-b}{a} = \frac {-7}{1}\) = -7; αβ = \(\frac {c}{a} = \frac {10}{1}\) = 10
\(\frac { \alpha }{ \beta } +\frac { \beta }{ \alpha } =\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 } }{ \alpha \beta } =\frac { { \left( \alpha +\beta \right) }^{ 2 }-2\alpha \beta }{ \alpha \beta } =\frac { 49-20 }{ 10 } =\frac { 29 }{ 10 } \)

Here t₂ - t₁ = t₃ - t₂
13 - 6 = 20 - 13 = 7
It is an Arithmetic series
Sum \(\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(a+l)\)
a = 6 : l = 97
a + (n - 1) d = 97
6 + (n - 1) (7) =  97
(n - 1) (7) = 97 - 6
(n - 1) (7) = 91
\(n-1=\frac{91}{7}=13\)
n = 13 + 1 = 14
Now \(S_{n}=\frac{14}{2}(6+97)=7 \times 103\)
6 + 13 + 20 +...+ 97 = 721

To find the coefficient of variation we need standard deviation
\(\sigma =\sqrt{\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}} \)
\(\frac{\Sigma x^{2}}{n} =\frac{765}{5}=153 \)
\(\left(\frac{\Sigma x}{n}\right)^{2} =(\bar{x})^{2}=6^{2}=36 \)
\(\sigma =\sqrt{(153)-36}=\sqrt{117} \)
\(=\sqrt{3 \times 3 \times 13} \)
\(\sigma =3 \sqrt{13} \)
Coefficient of variation \(=\frac{\sigma}{x} \times 100 \%\)
\(=\frac{3 \sqrt{13}}{6} \times 100 \%=\frac{\sqrt{13}}{2} \times 100 \%=\frac{3.60555}{2} \times 100 \% \)
\(=1.80277 \times 100 \%=180.277 \% \)
Coefficient of variation = 180.28 %

Sum of first k natural numbers  = \(\frac{k(k+1)}{2}=325\)
Sum of cube of first k natural numbers
\(=\left[\frac{k(k+1)}{2}\right]^{2} \)
\(=(325)^{2}=1,05,625 \)
\(1^{3}+2^{3}+3^{3}+\ldots+k^{3}=1,05,625 \)

Diameter of cylindrical pipe = 14 cm
Radius = 7 cm
Length of the pipe = Speed of the water
= 15 km = 15000 m
Length of the water tank = 50 m
Width of the water tank = 44 m
Height of the water tank = Water level
= 21 cm
= 0.21 cm
volume of water tank = l x b x h cu. units
= 50 x 44 x 0.21 = 462 m³
Volume of cylindrical Pipe = Volume of Rectangular tank
\(\frac{\pi r^{2} h}{} h =462 \)
\(\frac{22}{7} \times 0.07 \times 0.07 \times h =462 \)
\(\mathrm{h} =\frac{462 \times 7}{22 \times 0.07 \times 0.07} \)
\(=\frac{3234}{0.1078}=30000 \)  
Time required \(=\frac{30000}{15000}=2 \text { hrs. } \)

\( \frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1} =\tan ^{2} \theta \)
\(\text { LHS } =\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1} \)
\(=\frac{1-\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta}{\sin ^{2} \theta}-1} \)
\(=\frac{\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{\cos ^{2} \theta}}{\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{\sin ^{2} \theta}}\)
\(=\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{\cos ^{2} \theta} \times \frac{\sin ^{2} \theta}{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)} \)
\(=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\tan ^{2} \theta=\text { RHS }\)

In \(\triangle\)ABC, AD is the bisector of \(\angle\)A
Therefore by Angle Bisector of \(\angle\)A
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac{4}{3}=\frac{6}{A C}\) gives 4AC = 18. Hence, AC \(=\frac{9}{2}=4.5 \mathrm{~cm}\)

(- 6, 1) and (-3, 2)
The slope \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 2-1 }{ -3+6 } =\frac { 1 }{ 3 } \)

A x B = {(3,1), (3,7), (3,10), (4,1), (4,7), (4,10), (7,1), (7,7), (7,10), (8,1), (8,7), (8,10)}
We note that, R₁ ⊆  A x B. Thus, R₁ is a relation from A to B.

-1

x	-4	-3	-2	-1	0	1	2	3	4	5
x2	16	9	4	1	0	1	4	9	16	25
+x	-4	-3	-2	-1	0	1	2	3	4	5
y=x2+x	12	6	2	0	0	2	6	12	20	30

Draw the parabola by the plotting the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12), (4,20), (5, 30)

To solve: X² + 1 = 0, subtract X² + 1 = 0 from y = X² + x.

This is a straight line.
Draw the line y = x - 1.

x	-2	0	2
-1	-1	-1	-1
y	-3	-1	1

Plotting the points (-2, -3), (0, -1), (2, 1) we get a straight line. This line does not intersect the parabola. Therefore there is no real roots for the equation X² + 1 = 0.

Construction:
Steps (1) Draw a line segment BC = 5.6 cm
Steps (2) At B, draw BE such that \(\angle CBE={ 60 }^{ 0 }\)
Steps (3) At B draw BF such that \(\angle EBF={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector to BC, which intersects BF at O and BC at G.
Steps (5) With O as centre and OB as radius draw a circle
Steps (6) From B, marked an arc of 4 cm on BC at D.
Steps (7) The perpendicular bisector intersects the circle at I. Joined ID.
Steps (8) ID produced meets the circle at A. Now joined AB and AC. Then \(\triangle\)ABC is the required triangle.

Construction:
Step (1) Draw \(\bar { AB } =5.5cm\)
Step (2) Draw \(\angle BAE={ 25 }^{ 0 }\)
Step (3) Draw \(\angle FAE={ 90 }^{ 0 }\)
Step (4) Drawn the perpendicular bisector XY to AB which intersects AF at O and AB at G.
Step (5) With O as center and OA as radius drawn a circle
Step (6) XY intersects AB at G. On XY from G marked an arc at M such that GM = 4 cm
Step (7) Drawn PQ through M which is parallel to AB.
Step (8) PQ meets the circle at C and S.
Step (9) Joined AC and BC. Now \(\triangle\)ABC is the required triangle