Given that \(\angle SPT\) =  120°
\(\Rightarrow \angle OPS=\frac { { 120 }^{ ° } }{ 2 } =60^{ ° }\)
(as OP bisects \(\angle SPT\))
Also, \(\angle PTO=90^{ ° }\)
(as radius \(\bot\) tangent)
\(\therefore\)  In right triangle POS.
\(\cos { \angle OPS } =\frac { PS }{ OP } \)
\(\Rightarrow \frac { 1 }{ 2 } =\frac { PS }{ OP } \)
\(\Rightarrow OP=2PS\)

Given equation \(9x^{ 2 }-6ax+(a^{ 2 }+b^{ 2 })=0\)
\(x=\frac { 6a\pm \sqrt { (-6a)^{ 2 }-4\times 9\times (a^{ 2 }-b^{ 2 }) } }{ 2\times 9\\ } \)
\(\Rightarrow \quad x=\frac { 6a\pm \sqrt { 36a^{ 2 }-36a^{ 2 }-36b^{ 2 }) } }{ 18\\ } \)

\(\Rightarrow x=\frac { 6a+6b }{ 18 } ,x\frac { 6a-6b }{ 18 } \)
\(\Rightarrow x=\frac { 6a+b }{ 18 } ,x\frac { 6a-b }{ 18 } \\ \Rightarrow x=\frac { a+b }{ 3 } ,x=\frac { a-b }{ 3 } \)

3 : 2

Given In two concentric circles with common centre O, two chords of the outer circle which touch the inner circle at M and N.

To prove:  AB = CD
Construction: Join OM and ON.
Proof Since, OM and ON are radii of the inner circle through the points of contact M and N of the tangents AB and CD.
Also,  OM = ON   [radii of the inner circle] 
AB and CD are two chords of the outer circle which are equidistant from its centre O.
Hence, AB = CD

Let the same common difference of two APs be d. Given that, the first term of first AP and second AP are 2 and 7 respectively, then the APs are
 2, 2 + d, 2 + 2d, 2 + 3d,...
and 7, 7 + d, 7+ 2d, 7 + 3d,...
Now, 10th terms of first and second APs are 2 + 9d and 7 + 9d, respectively.
So, their difference
7 + 9d - (2 + 9d) = 5
Also, 21st terms of first and second Aps are 2 + 20d and 7 + 20d, respectively.
So, their difference is
7 + 20d -(2 + 20d) = 5
Also, if the \({a}_{n}\) and \({b}_{n}\) are the nth terms of first ans second AP.
Then,
\({b}_{n}-{a}_{n}=\left[7+\left(n-1\right)d\right]-\left[2+\left(n-1\right)d\right]=5\)
Hence, the difference between any two corresponding terms of such APs is the same as the difference between their first terms.

Let d be the common difference of an AP,
Given, \(a = \frac {1}{2}\) and \({a}_{8}=\frac{17}{6}\)
We know that,
\(a+(n-1)d={ a }_{ 8 }\)
\(\therefore\) \({ a }_{ 8 }=a+\left( 8-1 \right) d=\frac { 17 }{ 6 } \)
\(\Rightarrow\)  \(a+7d=\frac { 17 }{ 6 } \) 
\(\Rightarrow\)   \(\frac { 1 }{ 2 } +7d=\frac { 17 }{ 6 } \)     \(\left[ \because a=\frac { 1 }{ 2 } \right] \)
\(\Rightarrow\)    \(7d=\frac { 17 }{ 6 } -\frac { 1 }{ 2 } \)
\(\Rightarrow\)  \(7d=\frac { 17 }{ 6 } -\frac { 1 }{ 2 } \)
\(\Rightarrow\)  \(7d=\frac { 14 }{ 6 } \)
\(\Rightarrow\)  \(d=\frac { 1 }{ 3 } \)
Hence, the common difference is 1/3.

4.

Here, KAY is a tangent through one end A of a diameter Of a circle 

Also, CD || XAY ⇒ AB ⊥ CD
Since OM is perpendicular from centre O to the chord CD.
ஃ OM is perpendicular bisector of chord CD.
i.e.,  CM = MD = \(1\over2\)CD
Now, AM = 8 cm (given)
⇒ OM = AM - AO = 8 - 5 = 3 cm
In rt, \(\triangle\) OMC, we obtain
CM = \(\sqrt { { OC }^{ 2 }-{ OM }^{ 2 } } \)
= \(\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } \) = \(\sqrt{16}\) = 4 cm
Hence, CD = 2 CM = 2 x 4 = 8 cm.

30⁰

527.04km/h

Let AB be the tower of height h m, P and Q be the two points in line with the foot of the tower AB, such that
ㄥAPB=∝ and ㄥAQB=β
Consider rt. ㄥed ΔQBA,
\(\frac { AB }{ QB } \)=tan β
QB=AB.\(\frac { 1 }{ tan\beta } \)
=AB cot β....(i)
Again consider rt ㄥed ΔPBA,
\(\frac { AB }{ PB } \)=tan α
\(\frac { AB }{ PQ+QB } \)=tan α
⇒ PQ+QB=AB.\(\frac { 1 }{ tan\alpha } \)
=AB cot α....(ii)
Subtracting (i) from (ii), we obtain
PQ=AB cot α - AB cot β
PQ=AB (cot α - cot β)
PQ=h(cot α - cot β)

a = 80

\(60^{\circ}\)

Let AB is tower and car is at C on the highway.
 
In right \(\Delta\)ABC,
\(\frac { AB }{ BC } =\tan { { 30 }^{ o } } \)
\(\Rightarrow\)  \(\frac { 100 }{ BC } =\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow\)  \(BC=100\sqrt { 3 }\) m

AB'C' is the required triangle.

(i) Draw a circle of radius 3 cm with O as its centre
(ii) Draw AB as diameter of thc circle.
(iii) A point P is taken on outer circle and OP is joined.
(iv) Perpendicular bisector of Op is drawn interesting OP at Q.
(v) With Q as centre and OQ as radius a circle is drawn intersecting the smaller circle at A and B
(vi) PA and PB is joined
(vii) PA and PB are the required tangents.
Length of tangent = 4 cm.

Given quadratic equation  9x²-6x-2=0
Here a=9, b=-6, c=-2
D=b²-4ac
D=(-6)²-4X9X(-2)=36+72=108 > 0
Given quadratic has two unequal real roots.

Here, a = 2, d = 7 – 2 = 5 and n = 10.
We have a_n = a + (n – 1) d
So,  a₁₀ = 2 + (10 – 1) × 5 = 2 + 45 = 47
Therefore, the 10th term of the given AP is 47.

Given, equation is (x+2)³=2x(x²-1)
\(\Rightarrow\) x³ + 8 + 3x²(2) + 3x (2)² = 2x³ - 2x
[\(\because\) (a + b)³ = a³ + b³ + 3a²b + 3ab²]
\(\Rightarrow\) x³ + 8 + 6x² + 12x = 2x³ - 2x
\(\Rightarrow\) x³ + 8 + 6x² + 12x - 2x³ + 2x = 0
\(\Rightarrow\) -x³ + 6x² + 14x + 8 = 0
which is not of the form ax² + bx + c = 0,
because it has cubic term, i.e. x³.
\(\therefore\) It is not a quadratic equation.

Since, A(-1, y) and B(5, 7) lie on a circle with centre 0(2,-3y), Distance between OA and OB are equal
OA = OB
i.e., \(\sqrt { { \left( -1-2 \right) }^{ 2 }+{ \left( y+3y \right) }^{ 2 } } =\sqrt { { \left( 5-2 \right) }^{ 2 }+{ \left( 7+3y \right) }^{ 2 } } \)
\(\Rightarrow \quad \sqrt { 9+{ \left( 4y \right) }^{ 2 } } =\sqrt { { 3 }^{ 2 }+{ \left( 7+3y \right) }^{ 2 } } \)
Squaring on both sides, we get
9 + 16y² = 9y² + 42y + 58
\(\Rightarrow\)  y² - 6y - 7 = 0
\(\Rightarrow\)  (y + 1)(y - 7) = 0
\(\therefore\)  y = - 1, 7
Radius of circle, when y = - 1
(x - 2)² + (y - 3)² = r²
\(\because\)  (5, 7) lie on the circle,
\(\therefore\)  (5 - 2)² + (7 - 3)² = r²
\(\Rightarrow\)  9 + 16 = r²
\(\Rightarrow\)  r = 5 cm

By question  \(T_{ passenger }=\frac { 192 }{ x } \)

\(\Rightarrow \frac { 192 }{ x } -\frac { 192 }{ x+16 } =2\)
\(\Rightarrow 192(x+16)-192x=2(x^{ 2 }+16x)\)
\(\Rightarrow 192x+192\times 16-192x=2(x^{ 2 }+16x)\)
\(\Rightarrow x^{ 2 }+48x-32x-1536=0\)
\(\Rightarrow x(x+48)-32(x+48)=0\)
\(\Rightarrow x(x-32(x+48)=0\)
\(\Rightarrow (x-32)(x+48)=0\)
x=32 or -48
Since speed can't be negative, therefore - 48 is not possible.
:. Speed of passenger train = 32 km/!h % and Speed of fast train = 48 km/h.

Let AB = h km be the height of the mountain. Let C be a point at a distance of x km from the base of the mountain such that \(\angle\)ACB = 30\(°\) and let D be a point at a distance of 10 km from C along the same line. Then \(\angle\)ADB = 15\(°\) and AD = AC + DC = (x + 10) km

\(In\quad \Delta BAC,\quad tan\quad 30°=\frac { P }{ B } =\frac { AB }{ AC } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } } =\frac { h }{ x } \left[ \because tan 30°=\frac { 1 }{ \sqrt { 3 } } \right] \)
\(\Rightarrow x= h\sqrt { 3 } \)  ......(i)
\(In \Delta BAD, tan 15°=\frac { AB }{ AD } \)
\(\Rightarrow 0.27=\frac { h }{ x+10 } \left[ given, tan 15°=0.27 \right] \)
\(\Rightarrow\)  0.27(x + 10) = h  ......(ii)
On putting x = \(\sqrt { 3 } h\) from Eq. (i) in Eq. (ii), we get
\(0.27(\sqrt { 3 } h+10)=h\)
\(\Rightarrow 0.27\times \sqrt { 3 } h+0.27\times 10=h\)
\(\Rightarrow h(1-0.27\times \sqrt { 3 } )=0.27\times 10\)
\(\Rightarrow h(1-0.27\times 1.732)=2.7\)   \(\left[ \because \sqrt { 3 } =1.732 \right] \)
\(\Rightarrow\) h(1-0.47) = 2.7
\(\Rightarrow 0.53h=2.7\Rightarrow h=\frac { 2.7 }{ 0.53 } =5.09\approx 5km\)
Hence, the height of mountain is 5 km.

Given, diameter of circle = AB = 6 cm and centre is O.
Radius = OA = OB = 6/2 = 3 cm
  
Steps of construction
Step 1: Take a point O as centre and draw a circle of radius 3 cm
Step 2: Draw diameter AOB and take a point A on the circle
Step 3: Take OA as base and construct \(\angle OAT=90°\)at A
Step 4: Produce TA to T^' to get the required tangent TAT^' .Similarly, we can draw a tangent at point B or any point on the circle.

Yes, the new triangle is also equilateral.

\(\frac { 20 }{ \sqrt { 3 } } \)

Let n be the number of sides of the polygon.
We know that, sum of interior angles of a polygon of n sides
\(= ( 2n -4 ) \times { 90 } ^ { 0 }\)                                             ...(i)
Given, interior angles of polygon are in AP.
So, smallest angle = first term = a = \({ 120 } ^ { 0 }\)
and common difference, \(d = { 5 } ^ {0}\)
Then, sum of interior angles, \({ S }_{ n }=\frac { n }{ 2 } \left[ 2\left( { 120 }^{ 0 } \right) +\left( n-1 \right) { 5 }^{ 0 } \right] \)
 \(\left[ \because { S }_{ n }=\frac { n }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\} \right] \)
 \(={ \left\{ \frac { n }{ 2 } \left( 240+5n-5 \right) \right\} }^{ 0 }={ \left\{ \frac { n }{ 2 } \left( 5n+235 \right) \right\} }^{ 0 }\)     ...(ii)
From Eqs. (i) and (ii), we get
 \(={ \left\{ \frac { n }{ 2 } \left( 5n+235 \right) \right\} }^{ 0 }=\left( 2n-4 \right) \times { 90 }\)
\(\Rightarrow\)     \(5{ n }^{ 2 }+235b=\left( 2n-4 \right) \times 90\times \)    
\(\Rightarrow\)      \(5{ n }^{ 2 }+235n=360n-720\)
\(\Rightarrow\)      \(5{ n }^{ 2 }+235n=360n-720\)
\(\Rightarrow\)     \(5{ n }^{ 2 }-125n+720=0\)
\(\Rightarrow\)      \({ n }^{ 2 }-25n+144=0\)     [dividing by 5]
\(\Rightarrow\)       \({ n }^{ 2 }-9n-16n+144=0\)
\(\Rightarrow\)    \(n\left( n-9 \right) -16\left( n-9 \right) =0\)
\(\Rightarrow\)  \(\left( n-9 \right) -16\left( n-9 \right) =0\)
\(\Rightarrow\)       \(n=9,16\)
In n = 16, then the greatest angle, i.e.
 \({ a }_{ 16 }={ 120 }^{ 0 }+\left( 16-1 \right) { 5 }^{ 0 }={ 195 }^{ 0 }\)
which is not possible, since no interior angle of a polygon can be more than \({ 180 }^{ 0 }\)
\(\therefore\)     n = 9
Hence, nuumber of sides of the polygon is 9.

40 units

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.

Let two successive integral multiples of 5
be  x and x + 5.
ATQ x (x + 5) = 300 
\(\Rightarrow x^{ 2 }+5x-300=0\)
\(\Rightarrow (x+20)(x-15)=0\)
\(\Rightarrow \) x=-20 or x=15
When x=-20, multiples are -20 and -20+5= -15
When x=15 , multiples are 15 and 15+5 =20

-11 or 4

Let AB=b m be the height of the pole and AC=15 m be the length of the rope which is tied from top of the pole. 
In \(\Delta ABCsin { { 45 }^{ ° } } =\frac { AB }{ AC } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 2 } } =\frac { b }{ 15 } \)
\(\Rightarrow b=\frac { 15 }{ \sqrt { 2 } } \)
\(\Rightarrow b=\frac { 15 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } \)
\(\Rightarrow b=\frac { 15\times 1.414 }{ 2 } [\because \sqrt { 2 } =1.414]\)
\( \Rightarrow b=10.6m\)
Hence, the height of the pole is 10.6 m.
(ii) Applications of trigonometry.
(iii) Fun and entertainment.

As tangent radius are perpendicular to each other
∴ \(\angle \)OPT = 90°
Now, using Pythagoras theorem, 
PT² = OT² - OP²
⇒ PT² = (13)² - (5)²
PT² = 169 - 25 = 144
⇒ PT = 12 cm
In \(\triangle\)TAE and \(\triangle\)TBE 
\(\angle \)1 = \(\angle \)2
TE = TE     [common]
\(\angle \)TEA = \(\angle \)TEB
[each = 90° i.e., radius is ⊥ to tangent]
\(\triangle\)TAE = \(\triangle\)TBE   [by ASA congruence axiom]
ஃ AE = EB and TB = TA    [c.p.c.t.]
In \(\triangle\)AET
AE² = AT² - ET²
AE² = AT² - (8)²
⇒ x² = (12-x)² - 64
[ where, AE = x and AE = AP = x]
⇒ x² = 144 + x² - 24x - 64
⇒ 24 x = 144 - 64
⇒ x = \({80\over24}={10\over3}\)

94.64m

\(x = 1 , x = - {(a - b)^{2} \over (a + b)^{2}}\)

Given: In the given figure AB be a diameter. At is a tangent segment.

∴ ABT be a right-angled triangle.
Since AB = 5 cm, AT = 12 cm
∴ BT = \(\sqrt { { 5 }^{ 2 }+{ 12 }^{ 2 } } =13\quad cm\)
Construction: Join AC.
Since AB be a diameter. ∴ ACB = 90°
Here the right angled \(\triangle\)'s ABT and ACT are similar.
ஃ \(\frac { BT }{ AT } =\frac { AT }{ CT } \Rightarrow CT=\frac { { (AT) }^{ 2 } }{ BT } \)
= \(\frac { 12\times 12 }{ 13 } =\frac { 144 }{ 13 } cm\)

Ist instalment = Rs 1000
IInd instalment = Rs 1000 + Rs 100 = IIIrd instalment = Rs 1100 + Rs 100 = Rs 1200 and so on
Let number of instalments = n
\(\therefore\)  1000 + 1100 + 1200 + .... up to n terms = 118000
\(\Rightarrow\)  \({n\over2}[2\times1000+(n-1)100]=118000\)
\(\Rightarrow\)  n( 100n + 1900 ) = 236000
\(\Rightarrow\)  100n² + 1900n - 2360 = 0
\(\Rightarrow\)  n² + 19n - 2360 = 0
\(\Rightarrow\)  ( n + 59 ) ( n - 40 ) = 0.
\(\Rightarrow\) n = - 59 ( rejected ) or n = 40
\(\therefore\)   Total no. of instalment = 40th instalment
\(\therefore\)    a₄₀ = a + 39d
= 1000 + 39 x 100
= Rs 4900
Loan repaid i 30 instalments = \({30\over2}[2\times1000+(30-1)\times100]\)
\(={30\over2}(2000+2900)\)
= 15 x 4900 = Rs 73500
\(\therefore\)    Amount to be repaid after 30th instalment = 118000 - 73500 = Rs 44500 

Given A\(\Delta ABC\) , in which BC=5cm, \(\angle CAB=120°\) and \(\angle ABC=30°\) .
Then, \(\angle BCA=180°-30°-120°=30°\)

Since, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
i.e  ㄥAOB + ㄥCOD = 180^o
⇒  ㄥCOD =180^o  - ㄥAOB
⇒ ㄥCOD =  = 180^o  - 125^o = 55^o