Given, \(\sin { { \theta }_{ 1 } } +\sin { { \theta }_{ 2 } } +\sin { { \theta }_{ 3 } } =3\)
and \({ 0 }^{ 0 }<{ \theta }_{ 1 },{ \theta }_{ 2 },{ \theta }_{ 3 }\le { 90 }^{ 0 }\)
We know that, the maximum value of \(\sin { \theta } \quad is\quad 1\)
\(\therefore \quad \sin { { \theta }_{ 1 } } +\sin { { \theta }_{ 2 } } +\sin { { \theta }_{ 3 } } =3\)
\( \Rightarrow \sin { { \theta }_{ 1 } } =\sin { { \theta }_{ 2 } } =\sin { { \theta }_{ 3 } } =1\)
\(\Rightarrow { \theta }_{ 1 }={ \theta }_{ 2 }={ \theta }_{ 3 }={ 90 }^{ 0 }\)
\(Now,\quad \cos { { \theta }_{ 1 } } +\cos { { \theta }_{ 2 } } +\cos { { \theta }_{ 3 } } \)
\(=\cos { { 90 }^{ 0 } } +\cos { { 90 }^{ 0 } } +\cos { { 90 }^{ 0 } } \)
\(=0+0+0=0\) 

x²-x-56

We, have

Product of prime factors of 140
= 2 x 2 x 5 x 7= 2² x 5 x 7

Let E=Guessing the correct answer to a certain question
Probability (guessing the correct answer),
\(P(\overset { - }{ E } )=\frac { p }{ 12 } \)  [given]
Probability (not guessing the correct answer),
\(P(\overset { - }{ E } )=\frac { 3 }{ 4} \) [given]
We know that, \(1-P(E)=P(\overset { - }{ E } )\)
\(\Rightarrow \ 1-\frac { p }{ 12 } =\frac { 3 }{ 4 } =\frac { p }{ 12 } \\ \Rightarrow \ p=3\)
Hence, the value of p is 3.

\(50(3+\sqrt{3})m;\)To avoid accident everyone should follow road safety rules.

p = 3

Let the required ratio be k:1.
Then the coordinates of the point of division are \(\left( \frac { 2k+5 }{ k+1 } ,\frac { -3k+6 }{ k+1 } \right) \)
This point lies on the x-axis whose equation is y=0.
\(\therefore \ \frac { -3k+6 }{ k+1 } \ =\ 0\ \Rightarrow \ 3k=6,\ or\ k=2\)
 Line segment joining the two points is divided in the ratio 2:1 internally by x-axis. 

k = 3

Here, vessel is a combination of a hollow hemisphere and a hollow cylinder.

For cylindrical portion,
Diameter = AB = DC = 14 cm
\(\therefore\) Radius = OB = O'C = O' P
\(=\frac{A B}{2}=\frac{14}{2}=7 \mathrm{~cm}\)
Total length of vessel, PO = 13 cm
\(\therefore\) Length of cylinder, OO' = PO - O'P = 13 - 7 = 6 cm
For hemispherical portion,
Radius of hemisphere = Height of hemisphere = 7 cm
Now, the inner surface area of the vessel = Curved surface area of cylinder + Curved surface area of hemisphere
\(\begin{aligned} & =2 \pi r h+2 \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =2 \times \frac{22}{7} \times 7 \times 6+2 \times \frac{22}{7} \times(7)^2 \\ \end{aligned}\)
\(\begin{aligned} & =2 \times 22 \times 6+2 \times 22 \times 7 \\ \end{aligned}\)
\(\begin{aligned} & =44(6+7)=44 \times 13=572 \mathrm{~cm}^2 \end{aligned}\)

Given: AB, AC and AD are tangents. AB = 5 cm. 
TO find: AD 
Sol. AB and AC are tangents from the same point to the circle with centre O. 
⇒ AB = AC      .......(i)
(Length of the tangents from the same external point are equal). 
AC and AD are tangents from the same point to the circle with centre O. 
⇒ AC = AD    .....(ii)
(Length of the tangents from the same external point are equal) 
From (i) and (ii) 
∵ AB = AD = 5 cm 

(i) Use AA criterion to prove \(\Delta A B C \sim \Delta A D E\)
(ii) AE = \(\frac{15}{13}\) cm, DE = \(\frac{36}{13}\) cm
(iii) 30 cm²

Let a and d > 0 be the first term and common difference of an AP.
According to the given condition,
 \({a}_{1} + { a } _{ 5 } { = } { 26 }\)
\(\Rightarrow\)    a + a + (5 - 1)d = 26
\(\Rightarrow\)     2a + 4d = 26
\(\Rightarrow\)    a + 2d = 13     [dividing by 2]  ... (i)
Now,            \({a}_{2}\times a_4=160\)
\(\Rightarrow\)    \((a+d)\times(a+3d)=160\)
\(\Rightarrow\)  (13 - 2d + d) (13 - 2d + 3d) = 160   [from Eq. (i)]
\(\Rightarrow\)    (13 - d) (13 + d) = 160
\(\Rightarrow\)    \((13)^2-(d)^2=160\)
\(\Rightarrow\)    \({d}^{2}=169-160\)
\(\Rightarrow\)      \({d}_{2}=9\)
\(\Rightarrow\)       \(d = \pm 3\)
But d > 0
\(\therefore\)       d = 3
On putting d = 3 in Eq. (i), we get
   \(a + 2 \times 3 =13\)
\(\Rightarrow\)      a = 13 - 6
\(\therefore\)       a  = 7
Now, the sum of first seven terms,
  \({ S }_{ 7 }=\frac { 7 }{ 2 } \left[ 2\times a+\left( 7-1 \right) d \right] \)      
 \(=\frac { 7 }{ 2 } \left[ 2\times 7+6\times 3 \right] \)
 \(=\frac { 7 }{ 2 } \left[ 14+18 \right] \)
  \(=\frac { 7 }{ 2 } \left[ 14+18 \right] \) 

Let aeroplane be at B and let the two ships at C and D, such that their angles of depression from B are \(30°\)and\(60°\), respectively.Then, angles of elevation of D and C from B are \(30°\)and \(60°\)respectively.

We have, AB=1200 m
Here, one side AB is common in both triangles.
Let AC=x m and CD= y m
In \(\Delta BAC\) , we have
\(tan\quad 60°=\frac { A }{ B } \Rightarrow \sqrt { 3 } =\frac { 1200 }{ x } \)
⇒ \(x=\frac { 1200 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 1200\sqrt { 3 } }{ 3 } =400\sqrt { 3 } \)  ....(i)
In \(\Delta BAD\) , We have
\(tan\quad 30°=\frac { AB }{ AD }=\frac { AB }{ DC+CA }\)    [∵ AD=DC+CA]
On putting the value of x from Eq.(i) in Eq. (ii) we get
\(y=1200\sqrt { 3 } -400\sqrt { 3 } \)
⇒ y = 1800\(\sqrt { 3 } \)
⇒ y = 800 x 1.732 \(\left[ \because \sqrt { 3 } =1.732 \right] \)
\(\Rightarrow y=1385.6m\)
Hence, the distance between both ships is 1385. 6m.

(a) AD = 4 m
(b) Care towards nature and love for creature.

(a) When three coins are tossed simultaneously, then the number of possible outcomes = 8,
(i.e. , HHH, HTH, THH, TTH, HHT, HTT, THT, TTT)
Number of favourable outcomes (three heads) = l, (i.e., HHH)
\(\therefore\)Required probability = \(\frac{1}{8}\) 
(b) Number of favourable outcomes (exactly 2 3, (i.e., HTH, THH, HHT)
\(\therefore\)Required probability =\(\frac{3}{8}\) 
(c) Number of favourable outcomes (at least two heads) = 4, (i.e., HTH, THH, HHT, HHH)
Required probability =\(\frac{4}{8}=\frac{1}{2}\) 

Here, OABC is a square of side 7 cm 
⇒ OA = AB = BC CO = 7 cm
and OAPC is quadrant of a circle with centre O. 
Now, radius of the circle = 7 cm 
Hence, Area of shaded region 
= Area of square OABC - Area of quadrant OAPC 
= 7 x 7 - \(\frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \) x 7 x 7 x = \(\left( 49-\frac { 77 }{ 2 } \right) \) cm²
= \({98-77\over2}={21\over2}\) = 10.5 cm²

R(x,y) lies on the line segment joining the points P(a,b) and Q(b,a). Then P,Q and R lie on a line.
⇒ x(b-a)+a(a-y)+b(y-b)=0 ⇒ bx-ax+a²-ay+by-b²=0
⇒ b(x+y)-a(x+y)+(a²-b²) =0 ⇒ (b-a)(x+y)-(b-a)(b+a)=0
⇒ (b-a){(x+y)-(b+a)}=0 ⇒ (+y)=(b+a) [ Assuming a≠ b]

Given that coordinates of one end point of the diameter is (4, -1) and centre of the circle is (1,- 3).

Let coordinates of the other end of the diameter be (x, y). 
We know that the centre of the circle (1,-3) is the mid-point of diameter.
\(\Rightarrow \frac { 4+x }{ 2 } =1\) and \( \frac { (-1+y) }{ 2 } =-3\)
\(\Rightarrow 4+x=2\) and \( -1+y=-6\Rightarrow x=-2\) and \( y=-6+1=-5\)
Thus, coordinates of the other end of the diameter are (-2, -5).

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

Volume of milk = \(\frac {22}{7}\) x 1x 1x 4·2 = 13·2 m³
To booth I= 13·2 x \(\frac {3}{5}\) = 264 x3
= 7.92 m³
To booth II = 13·2 x \(\frac {2}{5}\) = 264 X 2
= 5.28 m³
Height in 1^st vessel = \(\frac {7.92}{3.96}\) = 2 m 
Height in 2^nd  vessel  \(=\frac { 5.28 }{ \frac { 22 }{ 7 } \times 1 } =\frac { 5.28\times 7 }{ 22 } \)
= 1.68 m.

Let length of the flag = x cm and breadth of the flag = y cm .
2x + 2y = 50
\(\Rightarrow\) x + Y = 25        ....(i)
(x - 3) (y + 2) = xy - 6
\(\Rightarrow\) xy + 2x - 3yx - 6 = xy - 6
\(\Rightarrow\) 2x - 3y = 0      .... (ii)
On solving the eqns. (i) and (ii),
x = 15 cm and y = 10 cm
\(\therefore\)  Length of the flag = 15 cm and Breadth of the flag = 10 cm
Significane: The saffron colour is symbol of courage and sacrifice

To find the class marks, the following relation is used.
\(x_{i}=\frac{\text { Upper class limit + Lower class limit }}{2}\)
Class size (h) for this data = 10
Taking 70 as assumed mean (a), d_i, u_i, and f_iu_i are calculated as follows.

From the table, we obtain
\(\sum f_{i}=35 \)
\(\sum f_{i} u_{i}=-2 \)
\(\text { Mean } \bar{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) \times h\)
\(=70+\left(-\frac{2}{35}\right) \times(10) \)
\(=70-\frac{20}{35} \)
\(=70-\frac{4}{7}\)
= 70 - 0.57
= 69.43
Therefore, mean literacy rate is 69.43%.

Given, cot \(\theta \) =\(\frac { 7 }{ 8 } \Rightarrow \frac { B }{ P } =\frac { 7 }{ 8 } \)
Let B = 7k and P = 8k
where, k is any positive integer.
Draw a right angled \(\Delta\)PQR, right angled at Q
In right angled \(\Delta\)PQR, H² = B² + P²
\(\Rightarrow\)  H² = (7k)² + (8k)²   [ by using pythagoras theorem]
\(\Rightarrow\)  H² = 49 k² + 64k² = 113k²
\(\Rightarrow\)  H = k\(\sqrt { 113 } \)    [taking positive square root since, side cannot be negative]
\(\therefore \quad \sin \theta=\frac{P}{H}=\frac{8 k}{k \sqrt{113}}=\frac{8}{\sqrt{113}}\)      

\(\text { and } \cos \theta=\frac{B}{H}=\frac{7 k}{k \sqrt{113}}=\frac{7}{\sqrt{113}}\)
\(\frac { (1+sin\theta )(1-sin\theta ) }{ (1+cos\theta )((1-cos\theta ) } \)=\(\frac { { 1 }^{ 2 }-{ sin }^{ 2 }\theta }{ { 1 }^{ 2 }-{ cos }^{ 2 }\theta } \)
\(\left[ \because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(=\frac { 1-{ \left( \frac { 8 }{ \sqrt { 113 } } \right) }^{ 2 } }{ 1-{ \left( \frac { 7 }{ \sqrt { 113 } } \right) }^{ 2 } } =\frac { 1-\frac { 64 }{ 113 } }{ 1-\frac { 49 }{ 113 } } =\frac { \frac { 113-64 }{ 113 } }{ \frac { 113-49 }{ 113 } } \frac { 49 }{ 64 } \) 

566.16 cm² , 408.8 m

Let us assume that the height of the tower AC be h m and ∠ABC=60^o , ∠ADC=30^o
As BC=4m
DC=9m   (given)

Consider a rt△ACB, we have
tan 60^o = \(\frac{AC}{BC}\)
⇒ \(\sqrt{3} = \frac{h}{4}\) --- (i)
Again, consider a rt. △ACD, we have
tan 30⁰ = \(\frac{AC}{DC}\)
⇒ \(\frac { 1 }{ \sqrt { 3 } } =\frac { h }{ 9 } \)  --- (ii)
Now, multiply (i) and (ii), we have
\(1=\frac{h}{4}\times\frac{h}{9}\)
⇒ 1= \(\frac {h^{2}} {36}\)
⇒ h²=36
⇒ h=\(\sqrt{36}\)=6 m
⇒ h=6m
Hence, height of the tower is 6m.

We know that the central angle of a circle is 360°.

\(\therefore\) Angled of each sector \(=\frac{360^{\circ}}{6}=60^{\circ}\)
\(\therefore\) \(\angle\)AOC = 60° and OA = OC   [radii of circle]
\(\therefore\) \(\angle\)OAC = \(\angle\)OCA = x (say)
[\(\because\) angles opposite to equal sides of a triangle are also equal]
In \(\Delta\)AOC, \(\angle\)AOC + \(\angle\)OAC + \(\angle\)OCA = 180°
[by angle sum property of triangle]
\(\Rightarrow\) 60° + x + x = 180° \(\Rightarrow\) 2x = 180° - 60°
\(\Rightarrow \quad x=\frac{120^{\circ}}{2} \Rightarrow x=60^{\circ}\)
\(\Rightarrow \quad \angle O A C=\angle O C A=60^{\circ}=\angle A O C\)
\(\therefore\) \(\Delta\)AOC is a equilateral triangle.
Area of \(\Delta\)AOC = \(=\frac{\sqrt{3}}{4} \times(28)^2=333.2 \mathrm{~cm}^2\)
[\(\because\)area of an equilateral triangle =\(\left.\frac{\sqrt{3}}{4}(\text { side })^2\right]\)
Now, area of sector OABCO  \(=\frac{\theta}{360^{\circ}} \times \pi r^2\)
\(=\frac{60^{\circ}}{360^{\circ}} \times \frac{22 \times(28)^2}{7}=\frac{22 \times 4 \times 28}{6}=410.67 \mathrm{~cm}^2\)
\(\therefore\) Area of segment ABCA
= Area of sector OABCO - Area of \(\Delta\)AOC
= 410.67 - 333.2 = 77.47 cm²
Now, area of six segments = 6 \(\times\) 77.47 = 464.82 cm²
Since, the cost of making the design is Rs 0.35 per cm².
\(\therefore\)  Total cost = 464.82 \(\times\) 0.35 = Rs 162.69

Let the speed while going be x km/h
Distance = 150 km 
\(\therefore \)  Time taken = \(\frac { 150 }{ x } \) hours
Now speed in return journey be (x + 10) km/h 
  \(\therefore \)Time taken in return journey 
\(=\frac { 150 }{ x+10 } \)
ATQ  \(=\frac { 150 }{ x } -\frac { 150 }{ x+10 } =2\frac { 1 }{ 2 } \)
\(\Rightarrow \frac { 150(x+10)-150x }{ x(x+10) } \frac { 5 }{ 2 } \)
\(\Rightarrow \frac { 1500 }{ x^{ 2 }+10x } =\frac { 5 }{ 2 } \)
\(\Rightarrow 2\times 1500=5(x^{ 2 }+10x)\)
\(\Rightarrow 600=x^{ 2 }+10x\)
\(\Rightarrow x^{ 2 }+10x-600=0\)
\(\Rightarrow (x+30)(x-20)=0\)
\(\Rightarrow x=-30\) (rejecting) x=20
Speed while going = 20 km/h  speed in return journey   and 20 + 10 = 30 km/h