Here \(\sqrt { 3 } \sin { \theta } -\cos { \theta } =0\) and \({ 0 }^{ ° }<\theta <{ 90 }^{ ° }\)
\(\Rightarrow \quad \sqrt { 3 } \sin { \theta } =\cos { \theta } \)
\(\Rightarrow \quad \frac { \sin { \theta } }{ \cos { \theta } } =\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \quad \tan { \theta } =\frac { 1 }{ \sqrt { 3 } } \)
\(=\tan { { 30 }^{ ° } } \left[ \because \quad \tan { \theta } =\frac { \sin { \theta } }{ \cos { \theta } } \right] \)
\(\therefore \quad \theta =\tan { { 30 }^{ ° } } \)

Here, \(\angle APB={ 50 }^{ ° }\)
\(\angle PAB=\angle PBA=\frac { { 180 }^{ ° }-{ 50 }^{ ° } }{ 2 } ={ 65 }^{ ° }\)
\(\angle OAB={ 90 }^{ ° }-\angle PAB\)
\(={ 90 }^{ ° }-{ 65 }^{ ° }={ 25 }^{ ° }\)

Here, \(a=\sqrt { 2 } ,a+d=\sqrt { 8 } =2\sqrt { 2 } \)
 \(d=2\sqrt { 2 } -\sqrt { 2 } =\sqrt { 2 } \\ \therefore Next\quad term=\sqrt { 32 } +\sqrt { 2 } \\ =4\sqrt { 2 } +\sqrt { 2 } \\ =5\sqrt { 2 } \\ =\sqrt { 50 } \)

Sum of zeroes = 6, Product of zeroes = 9
\(\therefore\)  Quadratic polynomial is x² - 6x + 9
Also x² - 6x + 9 = 0
\(\Rightarrow\)  (x - 3)(x - 3) = 0
\(\Rightarrow\)  x = 3, 3
Hence zeroes are 3, 3

Diameter of solid ball =Length of edge of cubical box = a
\(\therefore\) Volume of remaining space inside the box=Volume of cubical box-Volume of solid ball.
\(\sqrt { 3 } :\frac { a^{ 3 } }{ 6 } (6-\pi )\)

12

Here, point A lies on Y-axis, so its abscissa is zero and its ordinate is 3. So, the point A on Y: axis is A(0, 3).
Here, A=(x₁,y₁)=(0, 3) and B=(x₂,y₂)=(-5, 2)
\(\therefore \)AB =\(\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \)
 =\(\sqrt { { \left( -5-0 \right) }^{ 2 }+{ (2-3) }^{ 2 } } =\sqrt { { (-5) }^{ 2 }+{ (-1) }^{ 2 } } \)
=\(\sqrt { 25+1 } =\sqrt { 26 } \) units
Hence, the distance between A and B is \(\sqrt { 26 } \) units.

Here, total number of students in the class=23
Number of students from three houses A, B and C
=4+8+5=147
Number of students not from houses A, Band C=23-17=6
Hence, the required probability=\({6\over23}\)

15 cm

Here, Prabhsimran planted the treat at P₁(2,1), P₂(3,2) and P₃(5,4).
So, area of △ P₁P₂P₃
=\(\frac {1}{2}\)|2(2-4)+3(4-1)+5(1-2)|
=\(\frac {1}{2}\)|-4+9-5|=0
Thus, given points  P₁P₂P₃ are collinear or they form a straight line.
Now, Tavleen planted the trees at T₁(1,2), T₂(2,3) and T₃(3,3).
∴ Area of △ T₁T₂T₃
=\(\frac{1}{2}\)|1(3-3)+2(3-2)+3(2-3)|
=\(\frac{1}{2}\)|0+2-3|-\(\frac{1}{2}\)sq units
Thus, given points are not collinear or they do not form a straight line. Hence, Prabhsimran planted the trees in straight line. Planting trees help in making environment clean and green. So the two girls by planting trees are giving healthy environment to the society.

50^∘

Length of the arc=\(\frac{\theta\pi\times{r}}{180°}=\frac{120}{180}\times{\frac{22}{7}}\times{12}\)
=circumference of the base cone
Let radius of cone be r
⇒2xπxr=\(\frac{120}{180}\times{\frac{22}{7}}\times{12}\Rightarrow r=\frac{2}{3}\times{\frac{12}{2}}\)=4cm
r=4cm, l=12 cm
h²=128 ⇒ h=\(\sqrt{128}=8\sqrt{2}\)cm
Volume of the cone=\(\frac{1}{3}\times\pi\times r^{2}\times{h}=\frac{1}{3}\times{\frac{22}{7}}\times \left(4\right)^{2}\times{8}\times{\sqrt{2}}\)
=\(\frac{1}{3}\times{\frac{22}{7}}\times{16}\times{8}\times{1.414}\)cm³=189.61 cm³

45

BC'A' is the required triangle.

(i) Draw a circle of radius 4 cm with O as its centre
(ii) Draw AB as diameter of thc circle.
(iii) Take P and Q as twopointson extended diameter AB such that OP = 0Q = 6 cm.
(iv) Draw perpendicular bisector of OP and OQ intersecting Op and Oe at M and N respectively
(v) With M as centre and OM as radius draw a circle intersecting the 1st circle at T and S.
(vi) With N as centre and eN as radius intersecting the 1st circle at D and E
(vii) Join PT, PS, QD and QE
(viii) PT, PS, QD and QE are required tangents.

Given, a = -2, d = 0
The first four terms of the AP are -2, -2, -2, and -2.

sin² 30° cos² 45° + 4 tan² 30° + \(\frac{1}{2}\) sin² 90° - 2 cos² 90° + \(\frac{1}{24}\)
\(={ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\times \left( \frac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }+4\left( \frac { 1 }{ \sqrt { 3 } } \right) ^{ 2 }+\frac { 1 }{ 2 } { \left( 1 \right) }^{ 2 }-2\left( 0 \right) +\frac { 1 }{ 24 } \)
\(=\frac { 1 }{ 4 } \left( \frac { 1 }{ 2 } \right) +\frac { 4 }{ 3 } +\frac { 1 }{ 2 } +\frac { 1 }{ 24 } \)
\(=\frac { 1 }{ 8 } +\frac { 4 }{ 3 } +\frac { 1 }{ 24 } +\frac { 1 }{ 2 } \)
\(=\frac { 48 }{ 24 } =2\)

By substitution method
Given, pair of linear equations is
8x+5y=9  ...(i)
and 3x+2y=4   ...(ii)
Solve similar to Q.1 part(i)
Hence, x=-2 and y=5
By cross-multiplication method
Given, pair of linear equations is
2x+5y=9
\(\Rightarrow\) 8x+5y-9=0   ...(i)
and 3x+2y=4
\(\Rightarrow\) 3x+2y-4=0   
On comparing Eqs. (i) and (ii) with standard form of pair of linear equations, we get
a₁=8, b₁=5, c₂=-9
and a₂=3, b₂=2, c₂=-4
By using cross-multiplication method, we get
\(\frac { x }{ [(5)-4)-(2)(-9)] } =\frac { y }{ [(-9)(3)-9-4)(8)] } =\frac { 1 }{ [8\times 2-3\times 5] } \)
\(\left[ \because \quad \frac { x }{ { b }_{ 1 }{ c }_{ 2 }-{ b }_{ 2 }{ c }_{ 1 } } =\frac { y }{ { c }_{ 1 }{ a }_{ 2 }-{ c }_{ 2 }{ a }_{ 1 } } =\frac { 1 }{ { a }_{ 1 }{ b }_{ 2 }-{ a }_{ 2 }{ b }_{ 1 } } \right] \)
\(\Rightarrow \quad \frac { x }{ (-20+18) } =\frac { y }{ (-27+32) } =\frac { 1 }{ (16-15) } \)
\(\Rightarrow \quad \frac { x }{ -2 } =\frac { y }{ 5 } =\frac { 1 }{ 1 } \)
Now, take \(\frac { x }{ -2 } =\frac { 1 }{ 1 } \) and \(\frac { y }{ 5 } =\frac { 1 }{ 1 } \)
\(\Rightarrow\) x=-2 and y=5

(i) 36  
(ii) 42

\((i){1\over16}(ii){1\over16}(iii){1\over2}(iv){1\over4}(v){1\over25}(vi){3\over16}\)

535.71 m²

Let us assume that the height of the building be h m .
As CD=60 m
ㄥADB=30^o
ㄥCBD=60^o
Consider art . ΔBDC, we have
tan60^o=\(\frac { CD }{ BD } \)
⇒ \(\sqrt { 3 } =\frac { 60 }{ BD } \) [cross-multiply]
⇒ BD=\(\frac { 60 }{ \sqrt { 3 } } \).......(i)
Again, consider art . ΔABD, we have
tan30^o=\(\frac { AB }{ BD } \)
⇒ h=\(\frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ BD } \)       
⇒ h=\(\frac { 1 }{ \sqrt { 3 } } \times \frac { 60 }{ \sqrt { 3 } } \)   [from (i)]
⇒ h=\(\frac { 60 }{ 3 } \)=20 m
Hence, the height of the building is 20 m.

Let age of the boy be x years
\(\therefore \) Age of the father be (24 - x) years 
ATQ \(\frac { 1 }{ 4 } x(24-x)=x+9\)
\(\Rightarrow 24x-x^{ 2 }=4x+36\)
\(\Rightarrow x^{ 2 }-20x+36=0\)
\(\Rightarrow (x-18)(x-2)=0\Rightarrow x=18,x=2\)
\(\Rightarrow \) Age of the boy =2 years
and Age of Father = 24-2=22 years
If x = 18 then age of father =24- 18 = 6 (Not Possible)

Let DB is a tree and AD is the broken part of it which touches the ground at C.

Given: ㄥACB = 30\(\unicode{xb0} \) and BC = 8m
Let AB = x m and AD = y m
∴ Now, length of the tree = (x + y)m
In ΔABC,
\(\frac { AB }{ BC } \) = tan 30\(\unicode{xb0} \) ⇒ \(\frac { x }{ 8 } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \frac { 8 }{ \sqrt { 3 } } \)
and \(\frac { AB }{ AC } \)-sin 30\(\unicode{xb0} \) ⇒ \(\frac { x }{ y } =\frac { 1 }{ 2 } \)
⇒ y = 2x ⇒ y = 2 x \(\frac { 8 }{ \sqrt { 3 } } =\frac { 16 }{ \sqrt { 3 } } \)
Hence, total height of the tree
x + y = \(\frac { 8 }{ \sqrt { 3 } } +\frac { 16 }{ \sqrt { 3 } } =\frac { 24 }{ \sqrt { 3 } } =\frac { 24 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 24\sqrt { 3 } }{ 3 } \)= 8 x 1.732 = 13.856 m

Let A(a + b, b + a), B(a - b, b - a) and P(a, b) and P divides AB in k : 1, then :
\(a=\frac { k\left( a-b \right) +1\left( a+b \right) }{ k+1 } \)
\(\Rightarrow\)  a(k + 1) = k(a - b) + a + b
\(\Rightarrow\)  ak + a = ak - bk + a + b
\(\Rightarrow\)  bk = b
\(\Rightarrow\)  k = 1
\(\therefore\)   (a, b) divides A(a + b, b + a), B(a - b, b - a) in 1 : 1 internally.

Here, \(\frac { AB }{ DF } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } ,\frac { BC }{ EF } =\frac { 5 }{ 7.5 } =\frac { 2 }{ 3 } ,\frac { AC }{ DE } =\frac { 3 }{ 4.5 } =\frac { 2 }{ 3 } \)
As, \(\frac { AB }{ DF } =\frac { BC }{ EF } =\frac { AC }{ DE } \)
So, \(\triangle ABC\sim \triangle DFE\)  [by SSS similarity criterion]
Hence, figures (i) and (ii) are similar triangles, but no other pairs of triangles in the given figure are similar.

We know that, tangents drawn from an exterior point to a circle are equal in length.
AD = AF = x cm
BD = BE = y cm
CE = CF = z cm
Given, AB = 8 cm
⇒ AD + BD = 8cm
⇒ x + y = 8
BC = 10 cm ⇒ BE + CE = 10 cm
⇒ y + z = 10
and CA = 12 cm ⇒ CF + AF = 12 cm
⇒ z + x = 12
On adding Eqs. (i), (ii) and (iii), we get
2(x + y + z) =30
⇒ x + y + z = 15
On subtracting Eq. (ii) from Eq. (iv), we get
x =15 - 10 = 5
On subtracting Eq. (iii) from Eq. (iv), we get
y = 15-12 = 3
On subtracting Eq. (i) from Eq. (iv), we get
z=15-8=7
AD = xcm = 5 cm
BE= ycm = 3 cm
and CF = z cm = 7 cm
Hence, the length of AD, BE and CE are 5 cm, 3 cm and 7 cm, respectively.

(i)Total number of cards=52
Number of red coloured kings=2
p(a king of red colour)=\({2\over52}={1\over26}\)
(ii)Number of face cards=12
p(a face card)\(={12\over52}={3\over13}\)
(iii)Number of queen of diamonds=1
p(the queen of diamonds)\(={1\over52}\)

The area of rectangular sheet ABCD = 40 x 28 
= 1120 cm² 
The diameter of semicircle = 28 cm 
ஃ Radius of semicircle = 14 cm 
Area of semicircle = \(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times 14\times 14\quad { cm }^{ 2 }\)
= 308 cm² 
Thus, area of remaining paper = Area of paper - Area of semicircle 
= 1120 - 308 = 812 
∴ Required area = 812 cm² 

We know that, in natural numbers, the smallest and the largest number of three-digits which are multiples of 11 are 110 and 900 respectively. Therefore, the sequence of three-digit numbers which are multiples of 11 are 110, 121, 132, ..., 990. Thus, it is an A.P. with first term a = 110, common difference d = 9 and last term (a_n) = l = 990.
Let there are n terms in this sequence.
\(\therefore\)       a_n = 999
\(\Rightarrow\)   a + ( n - 1 )d = 999
\(\Rightarrow\)    110 + ( n - 1)11 = 999
\(\Rightarrow\)       11n = 990 - 99
\(\Rightarrow\)       11n = 891  \(\Rightarrow\) n = 81
Hence, the equired sum = \({81 \over 2}(110+990)\)
\(={81\times1100 \over 2}\)
= 81 x 550 = 44550

Let a man is standing on the deck of a ship at point A such that AB = 10 m and let CD be the hill.
Then, ∠EAD = 60° and ∠CAE = ∠BCA = 30° [alternate angles]
Let BC = x m = AE and DE = h m

In right angled ΔAED, 
\(\tan 60^{\circ}=\frac{P}{B}=\frac{D E}{E A}=\frac{h}{x}\)
In right angled ΔABC,
\(\tan 30^{\circ}=\frac{A B}{B C} \Rightarrow \frac{1}{\sqrt{3}}=\frac{10}{x}\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]\)
distance of the hill from the ship is 10\(\sqrt3\) m and height of the hill is 40 m.