\(\frac { \cos { { 45 }^{ ° } } }{ \sec { { 30 }^{ ° } } } +\frac { 1 }{ \sec { { 60 }^{ ° } } } =\frac { \frac { 1 }{ \sqrt { 2 } } }{ \frac { 2 }{ \sqrt { 3 } } } +\frac { 1 }{ 2 } \)
\(=\frac { 1 }{ \sqrt { 2 } } \times \frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } \)
\(=\frac { \sqrt { 6 } }{ 4 } +\frac { 1 }{ 2 } \)
\(=\frac { \sqrt { 6 } +2 }{ 4 } \)

On the diameter of a circle only two tangents can be drawn.

Here, a = 5, d = - 3
\(\because\)  l = a + (n -1)d
\(\because\)- 49 = 5 + (n -1)(- 3)
\(\Rightarrow\)-49 = 5-3n + 3
\(\Rightarrow\) 3n = 49 + 5 + 3
\(\Rightarrow\)n = 57/3 = 19^th term.

32

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.
\(\sqrt { 3 } :\frac { a^{ 3 } }{ 6 } (6-\pi )\)

1. Draw a line segment AC = 4.2 cm.
2. With A as a centre and radius 2.5 cm, draw two arcs, one above AC and one below AC.
3. With C as a centre and radius 3.5 cm, two arcs are drawn intersecting previous arcs at Band D.
4.Join AB, BC, AD and CD. Thus, ABCD is the required quadrilateral.
5. taking A as a centre and radius 6.3 cm, draw an arc' which intersects AC produced at C'.
6. Through C', draw C'B' and C'D' parallel to CB and CD, respectively.
Hence, AB'C'D' is the required quadrilateral similar to quadrilateral ABCD.

We know that, radius is perpendicular to the tangent at the point of contact.
\(\therefore \ \ \ OA\ \bot \ PA\\ \Rightarrow \ \angle OAP={ 90 }^{ \circ }\)

In , \(\triangle OAP\)
 PO² = PA² + AO²     [by Pythagoras theorem]
\(\Rightarrow \) (17)² = (PA)² + (8)²
\(\Rightarrow \) (PA)² = 289 - 64 = 225
\(\Rightarrow \) PA = \(\sqrt { 225 } \) = 15 cm
Hence, the length of the tangent from point P is 15 cm.

Slant height of each cone = \(\sqrt{{8}^{2}+{15}^{2}}\)
\(=\sqrt{64+225}=\sqrt{289}\)
= 17 cm

Total surface area of the shape so formed 
= 2 X Curved surface area of cone|
\(=2\times{{22}\over{7}}\times8\times17=854.86\) cm²

90^o

\(\left| PQ \right| =\sqrt { { (4-2) }^{ 2 }+{ (1-3) }^{ 2 } } =\sqrt { { 2 }^{ 2 }+{ (-2) }^{ 2 } } \) 
\( =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } units\)

Given: equilateral triangle of side a units. As B is at origin, therefore, coordinates of B are (O, 0). As BC = a and C lies on x-axis, coordinates Of C are (a, O). 
Let coordinates of A be (x,y)

As, AB=BC=AC
\(\sqrt { { (x-0) }^{ 2 }+{ (y-0) }^{ 2 } } =a=\sqrt { { (x-a) }^{ 2 }+{ (y-0) }^{ 2 } } \)
Squaring, we get
\({ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }={ x }^{ 2 }+{ a }^{ 2 }-2ax+{ y }^{ 2 }\)   ...(i)
\(\Rightarrow \ { x }^{ 2 }+{ y }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+{ a }^{ 2 }-2ax\)
\(\Rightarrow \ 2ax-{ a }^{ 2 }=0\ \Rightarrow a(2x-a)=0\)
\(\therefore \ \ x=\frac { a }{ 2 } \)
Substituting \(x=\frac { a }{ 2 } \)in (i), we get
\({ \left( \frac { a }{ 2 } \right) }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\)
\(\Rightarrow \ { y }^{ 2 }={ a }^{ 2 }-\frac { { a }^{ 2 } }{ 4 } =\frac { 3{ a }^{ 2 } }{ 4 } \)
 \(\Rightarrow \ y=\frac { \sqrt { 3a } }{ 2 } \)
∴ Coordinates of A are \(\left( \frac { a }{ 2 } ,\frac { \sqrt { 3a } }{ 2 } \right) \)
∴ Coordinates of vertices are A  \(\left( \frac { a }{ 2 } ,\frac { \sqrt { 3a } }{ 2 } \right) \) , B(0,0) and C(a,0)

AB is diameter, AC=24cm, BC=10cm
and ㄥACB=90⁰ [Angle insemicircle]
\(∴\ AB=\sqrt{(24)^2+(10)^2}cm\)
\(=\sqrt{576+100}cm=\sqrt{676}cm\)
=26cm
∴ Area of shaded portion = area of semicircle - area of ΔACB
\(=\left[{1\over 2}\times{22\over 7}\times169-120\right]cm^2\)
=[265.57-120]cm²=145.57cm²

Probability of winning a game = 0.3
\(\therefore\) Probability of losing the game = 1 - Probability of winning the game = 1 - 0.3 = 0.7

a₁=2x1+1=3
a₂=2x2+1=5
a₃=2x3+a=7
sum=3+5+7=15

 \(\sqrt{2},\sqrt{8},\sqrt{18}......\) =  \(\sqrt{2},2\sqrt{2},3\sqrt{2}.....\)
Next term is \(4\sqrt{2}\)

Let side of bigger square be x m and side of smaller square be y m
ATQ x²+y² = 640   ...(i)
and 4x-4y=64
\(\Rightarrow x-y=16\Rightarrow x=16+y\)
Substituting the value of x in equation (i) we get 
(16+y²)+y²=640
\(\Rightarrow 256+y^{ 2 }+32y+y^{ 2 }=640\)
\(\Rightarrow 2y^{ 2 }+32y-384=0\)
\(\Rightarrow y^{ 2 }+16y-192=0\)
\(\Rightarrow (y+24)(y-8)\)
\(\Rightarrow \) y=8 or y=-24 (Rejecting)
When y=8, then x=16+8=24 
\(\therefore \) Sides of squares are 8m and 24 m

Let AB and CD be the two posts such that AB = 2 CD. Let M be the mid-point of CA. Let \(\angle CMD=\theta \) and \(\angle AMD=90°-\theta \)
Clearly, CM=MA=\(\frac{1}{2}\)k
Let CD = h. then AB = 2h
Now, \(\frac{AB}{AM}\)=tan(\(90°-\theta \))= cot \(\theta\)
\(\frac { h }{ \left( \frac { k }{ 2 } \right) } =cot\theta \)
cot \(\theta\)=\(\frac{4h}{k}\)  ...(i)
Also in \(\Delta CMD\), \(\frac{CD}{CM}\)=tan \(\theta\)
\(\frac { h }{ \left( \frac { k }{ 2 } \right) } =tan\theta \)
tan \(\theta\)=\(\frac{2h}{k}\) ...(ii)
Multiplying (i) and (ii) \(\frac{4h}{k}\) x  \(\frac{2h}{k}\)=1
h²=\(\frac { { k }^{ 2 } }{ 8 } \)
h=\(\frac { k }{ 2\sqrt { 2 } } =\frac { k\sqrt { 2 } }{ 4 } \)

By elimination method:
Given, x + y = 5 .....(i)
and 2x - 3y = 4 ......(ii)
On multiplying eqn. (i) by 3 and eqn. (ii) by 1 and then adding them, we get
3(x + y) + 1(2x - 3y) = 3 \(\times\) 5 + 1\(\times\)  4
\(\Rightarrow\) 3x + 3y + 2x - 3y = 15 + 4
\(\Rightarrow\) 5x = 19
\(\therefore \quad x=\frac { 19 }{ 5 } \)
On putting \(x=\frac { 19 }{ 5 }\) in Eq. (i), we get 
\(\frac { 19 }{ 5 } +y=5\)
\(\Rightarrow \quad y=5-\frac { 19 }{ 5 } \)
\(\Rightarrow \quad y=\frac { 25-19 }{ 5 } \)
\(\therefore \quad y=\frac { 6 }{ 5 } \)
Hence, \(x=\frac { 19 }{ 5 } \) and \(y=\frac { 6 }{ 5 } \)
By substitution method:
Given,  x + y = 5          .....(i)
and 2x - 3y = 4            .....(ii)
From eq. (i), we have y = 5 - x .....(iii)
On substituting y from eqn. (iii) in eqn. (ii),
2x - 3 (5 - x) = 4
\(\Rightarrow\) 2x - 15 + 3x = 4
\(\Rightarrow\) 5x = 19
\(\Rightarrow \quad x=\frac { 19 }{ 5 } \)
On substituting x=\(\frac { 19 }{ 5 } \) in eqn. (iii), we get
\(y=5-\frac { 19 }{ 5 } \)
\(\Rightarrow \quad y=\frac { 6 }{ 5 } \)
Hence,\(\ x=\frac { 19 }{ 5 } \) and \(\ y=\frac { 6 }{ 5 } \)

Since, 237 > SI
On applying Euclid's division algorithm, we get
237 = 81 x 2 + 75    ....(i)
81 = 75 x 1 + 6         ...(ii)
75 = 6 x 12 + 3         .. (iii)
6 = 3 x 2 + 0             ...(iv)
Hence, and HCF (81, 237) = 3
In order to write 3 in the form of 81x + 237y, we move backwards as follows:
3 = 75 - 6 x 12                 [From (iii)]
= 75 - (81 - 75 x 1) x 12  [Replace 6 from (ii)]
= 75 - (81 x 12 - 75 x 1 x 12)
= 75 - 81 x 12 + 75 x 12
= 75 + 75 x 12 - 81 x 12
= 75 ( 1 + 12) - 81 x 12
=75 x 13 - 81 x 12
= 13(237 - 81 x 2) - 81 x 12   [Replace 75 from (i)]
= l3 x 237 - 81 x 2 x l3 - 81 x 12
= 237 x 13 - 81 (26 + 12)
= 237 x 13 - 81 x 38
= 81 x (-38) + 237 x (13)
= 81x + 237y
∴ Hence x = - 38 and y= 13
Note that the values of x and yare not unique.

Let y = x + 5, Then, the graph of y = x + 5 is a straight line which interscts the X-axis at x = -5, so -5 is the zero of the given polynomial.

15.84 seconds

94.64m

Total number of pieces = 8 + 10 = 18
(i) No.of triangles = 8. Hence, P(triangle is lost) = \(\frac{8}{18}=\frac{4}{9}\)
(ii) No.of squares = 10. Hence, P(square is lost) = \(\frac{10}{18}=\frac{5}{9}\)
(iii) No.of squares of blue colour = 6. So, P(square of blue colour is lost) = \(\frac{6}{18}=\frac{1}{3}\)
(iv) No.of triangles of red colour = 8 - 3 = 5. So, P(triangle of red colour is lost) = \(\frac{5}{18}\)

Let CD = h m be the height of the tower. At point D of the tower, a man is standing and observes the car at an angle of depression of 30°. After six seconds, the angle of depression of the car is 60°.
i.e. \(\angle\)ODA = 30° and \(\angle\)ODB = 60°
\(\Rightarrow\) \(\angle\)DAC = \(\angle\)ODA = 30°  [alternate angles]
and \(\angle\)DBC = \(\angle\)ODB = 60°  [alternate angles]
Let AB = y m and BC = x m
In right angled \(\Delta\)BCD,

\(\begin{array}{rlrl} \tan 60^{\circ} & =\frac{P}{B}=\frac{C D}{B C} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad \sqrt{3} & =\frac{h}{x} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad h & =\sqrt{3} x & {\left[\because \tan 60^{\circ}=\sqrt{3}\right]} \end{array}\)
\(\Rightarrow \quad h = \sqrt3 x\)....(i)
In right angled \(\Delta\)ACD,
\(\begin{array}{rlrl} \tan 30^{\circ} & =\frac{C D}{A C}=\frac{C D}{A B+B C} & & {[\because A C=A B+B C]} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{h}{x+y} & & {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad x+y & =h \sqrt{3} & \end{array}\)
\(\Rightarrow \quad x + y=\sqrt3 x(\sqrt3)\) [from Eq. (i)]
\(\Rightarrow\) x + y = 3x  .....(ii)
It is given that a car moves from point A to B in six seconds. Let its speed be k km/s.
\(\therefore \quad \text { Time }=\frac{\text { Distance }}{\text { Speed }}\)
\(\Rightarrow \quad 6=\frac{y}{k} \Rightarrow y=6 k\)
On putting y = 6k in Eq. (ii), we get
x + 6k = 3x \(\Rightarrow\) 6k - 2x \(\Rightarrow\) x = 3k
\(\therefore \quad \text { Time }=\frac{\text { Distance }}{\text { Speed }}=\frac{x}{k}=\frac{3 k}{k}=3 \mathrm{~s}\)
Hence, the car moves from point B to point C in 3s.

Given, \(\triangle ABC\sim \triangle EDF\)
Also, AB = 5 cm, AC = 7 cm, DF = 15 cm
and DE = 12 cm                         .... (i)
Since, \(\triangle ABC\sim \triangle EDF\)
\(\therefore \frac { AB }{ ED } =\frac { AC }{ EF } =\frac { BC }{ DF } \)
[ ∵ Corresponding sides of similar triangles are proportions]
\(\Rightarrow \frac { 5 }{ 12 } =\frac { 7 }{ EF } =\frac { BC }{ 15 } \)  [from Eq. (i)]

On taking first and second terms, we get
\(\frac { 5 }{ 12 } =\frac { 7 }{ EF } \Rightarrow EF=\frac { 7\times 12 }{ 5 } \) = 16.8 cm
On taking first and third terms, we get
\(\frac { 5 }{ 12 } =\frac { BC }{ 15 } \Rightarrow BC=\frac { 5\times 15 }{ 12 } \) = 6.25 cm
Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm.

1268 m

44 cm

Number of blue marbles = 35
Number of white marbles = 25
Number of red marbles = 40
Total marbles in box = 35 + 25 + 40
=100
(i) Probability of a white marble = P (white) 
=\(\frac{25}{100}=\frac{1}{4}\)
(ii) Number of non-blue marbles = 100-35
 = 65
P (not blue) = \(\frac{65}{100}=\frac{13}{20}\) 
(iii) Number of white or blue marbles= 25 + 35
= 60
Number of neither white nor blue marbles = 100-60
= 40
P (neither white nor-blue marbles) =\(\frac{40}{100}=\frac{2}{5}\)

Let OP intersect AB at a point C. Here, PA and PB are the two tangents from a point P lying outside the circle, to the circle with centre O. 

\(\angle \)APO = \(\angle \)BPO [∵ O lies on the bisector of ZAPB] 
Now, in \(\triangle\)ACP and \(\triangle\)BCP, we have 
AP = BP [tangents from an external pointl 
PC = PC [commonl]
\(\angle \)APO = \(\angle \)BPO [proved above] 
⇒ \(\triangle\)ACP ≅ \(\triangle\)BCP   [by SAS congruence axiom]
⇒ AC = BC   [c.p.c.t.] 
and  \(\angle \)ACP = \(\angle \)BCP 
= \(1\over2\) x 180° = 90° [c.p.c.t.]
Hence, OP is the perpendicular bisector of AB. 

Given: Two concentric circles of radii 3 cm and 5 cm. 
Required: A tangent from any point on the outer circle to inner circle. 
Steps of Construction :
Draw two concentric circles with radii 3 cm and 5 cm, mark O as their center. 
Take any point P on the outer circle join OP. 
Draw AB, the perpendicular bisector of OP, let it intersect OP in M. 
With M as centre and radius MP or MO, draw a semi-circle and let it intersect the inner circle in T.

Join PT.
Thus, PT is the required tangent.
On measurement, PT 4 cm
By calculation, PT = \(\sqrt { { OP }^{ 2 }-{ OT }^{ 2 } } \) 
= \(\sqrt { 25-9 } =\sqrt { 16 } \)
= 4 cm.

Here, first term (a) of an A.P. is -5 and the last term (a_n) is 45.
Let d be the common difference and n be the number of terms.
\(\therefore\)             a + ( n - 1 )d = a_n
\(\Rightarrow\)   - 5 + ( n - 1 )d  = 45
\(\Rightarrow\)    ( n - 1 )d = 50        ...(i)
Also, S_n = 120
\(\Rightarrow\)         \({n\over2}(a+l)=120\)
\(\Rightarrow\)    n ( - 5 + 45 ) = 240
\(\Rightarrow\)    n ( 40 ) = 240
\(\Rightarrow\)    n = 6
From (i), we have|
( 6 - 1 )d = 50
\(\Rightarrow\)   5d = 50
\(\Rightarrow\)    d = 10
Hence, the required number of terms is 6 and the common difference is 10.