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If the angle between two radii of a circle is 130° , then find the angle between the tangents at the ends of radii?

Answers:

In the above given figure
Angle T +. Angle. O=180 (ptqo cyclic quad.)
AngleT=180130
Angle T=50
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Ans.
O is the centre of the circle.
Given, <POQ = 130º
PT and QT are tangents drawn from external point T to the circle
<OPT = <OQT = 90º [ Radius is perpendicular to the tangent at point of contact]
In quadrilateral OPTQ,
<PTQ + <OPT + <OQT + <POQ = 360º
=> <PTQ + 90º + 90º + 130º = 360º
=> <PTQ = 360º – 310º = 50º
Thus, the angle between the tangents is 50º.
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If ratio of the roots of equation px^{2}+qx+q=0 is a:b, prove that √a÷b + √b÷a + √q÷p=0.

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Let the roots of the given equation be {tex}\alpha {/tex} and {tex}\beta{/tex}.
Then, {tex}\alpha {/tex} + {tex}\beta{/tex} = {tex}{{  q} \over p}{/tex} and {tex}\alpha .\beta = {q \over p}{/tex}
Given: {tex}{\alpha \over \beta } = {a \over b}{/tex}
Now, {tex}\sqrt {{a \over b}} + \sqrt {{b \over a}} + \sqrt {{q \over p}} = 0{/tex}
=> {tex}\sqrt {{\alpha \over \beta }} + \sqrt {{\beta \over \alpha }} + \sqrt {\alpha \beta } = 0{/tex} [Since, {tex}{\alpha \over \beta } = {a \over b}{/tex} and {tex}\alpha .\beta = {q \over p}{/tex}]
=> {tex}{{\sqrt \alpha } \over {\sqrt \beta }} + {{\sqrt \beta } \over {\sqrt \alpha }} + \sqrt {\alpha \beta } = 0{/tex}
=> {tex}{{\alpha + \beta + \alpha \beta } \over {\sqrt {\alpha \beta } }} = 0{/tex}
=> {tex}\alpha + \beta + \alpha \beta = 0{/tex}
=> {tex}{{  q} \over p} + {q \over p} = 0{/tex} [Since, {tex}\alpha + \beta = {{  q} \over p}{/tex} and {tex}\alpha \beta = {q \over p}{/tex}]
=> {tex}q+q=0{/tex}
=> 0 = 0
Hence proved.
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In quadratic equation ax^{2} + bx + c =0. a b c are in AP and all are positive .The root of equation p and q which is an integer . Find the value of p+q+pq

Answers:

let a = ed
b = e
c = e+d
therefore, (ed)x^{2} + (e)x +(e+d) =0
therefore p+q = ed/e and pq = e+d/e
p + q + pq = e  d +e +d /e
= 2e/e
= 2
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A trader bought a number of articles for Rs.900. Five articles were found damaged. He sold
each of the remaining articles at Rs.80 in the whole transaction. Find the number of articles he
bought.

basu
plz try to answer my non detail question above

don worry fr such questions they wont come any way

Not easy to understand
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Q.ABC is a triangle in which <B = 90°.A circle is drawn with AB as the diameter intersecting the hypotenuse AC at P.Prove that the tangent drawn to the circle at P, bisect BC.

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Given: triangle ABC is a right angled triangle, right angled at B, angle ABC = 90^{o}
A circle is drawn, taking AB as diameter, intersects AC at P. PQ is a tangent, intersects at Q.
To prove: BQ = QC
Construction: Join BP.
Proof: PQ = BQ ..........(i) [Tangents drawn from an external point are equal]
Therefore, angle PBQ = angle BPQ [Angles opposite to equal side]
Also, angle APB = 90^{o} [Angle of semicircle]
And angle APB + angle BPC = 180^{o} [Linear pair]
Now, angle BPC = 180^{o}  90^{o }= 90^{o}
Since, angle BPC + angle PBC + angle PCB = 180^{o} [Angle sum property of a triangle]
=> angle PBC + angle PCB = 180^{o}  angle BPC = 180^{<font size="2">o</font>}  90^{<font size="2">o </font>}= 90<font size="2">^{o}</font><font size="2"> .........(ii)</font>
Now, angle BPC = 90^{o}
=> angle BPQ + angle CPQ = 90^{o} ..........(iii)
From eq. (i) and (ii), we get,
angle PBC + angle PCB = angle BPQ + angle CPQ
=> angle PCQ = angle CPQ [Since angle BPQ = angle PBQ]
=> PQ = QC [In triangle PQC, opposite sides of equal angles] ..........(iv)
From, eq. (i) and (iv), we get,
BQ = QC
Hence proved.
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A tarder bought number of article for Rs. 900, five article are found to be damage, he sold remaining article for Rs. 80.find the number of article.

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is_/3,_/6_/9_/12 form an a.p. ??
sir I need answer immediately

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Hi Sir,
When will you update class 10 syllabus for academic year 20172018 and corresponding chapter weightages and clear understanding of the board exam pattern change?.
It will be grateful if you could update it soon as from April many schools would start the syllabus for 20172018

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The sum of n term of an ap is zero show that the sum of n term is  am(m+n) /n1 a being the first term.

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Oneforth of a herd of a camels was seen in the forest.Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river.Find the total number of camels.

Answers:

Ans. Let total number of camel = x^{2}
Nunber of camel seen in the forest = {tex}x^2\over 4{/tex}
Number of camel gone to mountain = {tex}2\sqrt {x^2} = 2x {/tex}
Number of camel seen on river bank = 15
According To Question,
{tex}=>{ x^2\over 4 } + 2x + 15 = x^2{/tex}
{tex}=> x^2 + 8x + 60 = 4x^2{/tex}
{tex}=> 3x^2 8x 60= 0{/tex}
=> 3x^{2}  18x + 10x  60 = 0
=> 3x(x6) +10(x6) =0
=> (x6)(3x+10) = 0
=> x = 6, 10/3
x cannot be ve.
So number of camel = 6^{2} = 36
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What constant must be subtracted or added from equation 4x²+12x+8=0 to solve it by method of completing square.

Answers:

ADD 1 AND SUBTRACT 1
(2X+3)^{2 }1=0
A^{2 }B^{2} FORMULA
2X+3+1=0 2X+31=0
2X+4=0 2X+2=0
2X=4 2X=2
X=2 X=1
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if one root of the quadratic equation is 3+2root5/4.then find the other root

Plz send answer. I have my maths exam tomorrow
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Find four consecutive terms in AP whose sun is 20 and the sum of whose square is 120 ?

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Ans. Let first term of A.P. is a2d
Common difference is d,
Then four terms are, a2d, ad, a and a+d
According To Ques,
=> a  2d + a  d + a + a + d = 20
=> 4a  2d = 20
=> 2a  d = 10
=> d = 2a  10 ……………(1)
Also,
=> (a2d)^{2} + (ad)^{2} + a^{2} +(a+d)^{2} = 120
put value of d from (1)
=> (a  4a+20)^{2 }+(a2a+10)^{2 }+ a^{2} +(a+2a10)^{2}= 120
=> (203a)^{2} + (10a) + a^{2} + (3a10)^{2 }= 120
=> 400 + 9a^{2}  120a + 100+ a^{2} 20a + a^{2}+ 9a^{2 }+ 100  60a = 120
=> 20a^{2}  200a + 600 = 120
divide by 20
=> a^{2}  10a + 30= 6
=> a^{2}  10a + 24 = 0
=> a^{2 } 6a  4a + 24 =0
=> a(a6) 4(a6) = 0
=> (a6)(a4) = 0
=> a = 6, 4
if a = 6 then d = 2
and if a = 4 then d = 2
First A.P = 2, 4, 6, 8,....
second A.P. = 8, 6, 4, 2,...
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Two taps running together can fill a cistern in 6 minutes .if one tap takes 5 minutes more than other to fill it , find the time in which each tap fill the cistern separeately?

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Let the first tap take x minutes to fill the cistern, then second tap will take (x+5) minutes to fill it.
The first tap will fill the 1/x of the cistern in 1 minute and second tap will fill 1/(x+5).
Together, they will fill 1/6.
So, equation becomes
{tex}{1\over x}{/tex}+ {tex}{1\over x+5}{/tex}= {tex}{1\over 6}{/tex}
x^{2}+5x= 12x+30
x^{2}7x30 =0
x^{2}10x+3x30=0
x(x10)+3(x10)=0
x= 10
Therefore, first tap will fill in 10 minutes alone and the second tap will take 15 minutes to fill the cistern.
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Find an ap whose first term is 1 and the sum of first four terms of one third of the sum of next four terms .

Answers:

a=1
S_{4}= {tex}{1\over 3}{/tex}[S_{8}S_{4}]
4\2 [ 2+3d] = 1\3 [2+7d23d]
2 (2+3d) = 1/3 (4d)
6(2+3d) = 4d
12 +18d = 4d
12 = 14d
6/7 =d
So, A.P. becomes 1,1/7,5/7....
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Determine the ratio in which the point z(a,5) divides the line segment joining the points X(2,1) and y(5,8).also find the value of a?

5 = 5m1 + 2m2 / m1 + m2
5m1  5m2 = 5m1 + 2m2
10m1 = 7m2
m1:m2 = 7:10
The ratio is 7 : 10
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Chance of winning a game are 60%.if Anil has played the game 20 times ,how many times he can expect to loose?

Answers:

Chance of losing the game = 40%
Total no. of games played = 20
No. of times lost = 40% of 20 = 8 times
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The mid point of the line segment joining (2a,4) and (2,2b) is (1,2a+1).find the value of b?

Answers:

Midpoint of two given points are
{(x_{1} + x_{2})/2, (y_{1} + y_{2})/2}
According to question,
(2a  2)/2 = 1
2a  2 = 2
2a = 4
a = 2
Again, (4 + 2b)/2 = 2a + 1
4 + 2b = 4a + 2
2b = 4 × 2 + 2  4
2b = 8  2
b = 3
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Mid point = (x_{1}+x_{2})/2 ; (y_{1}+y_{2})/2
1 = (2a+1)/2
2 = 2a+1
2a = 1
a =1/2
2a+ 1 = (4+2b)/2
4a+2 = 4 +2b
4a2 = 2b
2a1 = b
1  1 =b
b=0
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If one root of the quadratic equation is 3+2root5/4 then what will be the other root

Plz reply
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Out of a number of Saras birds ,one fourth of the number are moving in lotus plants ,one ninth coupled alongwith one fourth as well as seven times the square root of the number move on a hill . 56 birds remain in vakula trees .what is the total number of birds?

Please send answer for this question urgently.tomorrow is my maths exam
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Answers:

Ans. Let the number of birds be x^{2} Number of birds moving about lotus plant ={tex}x^2\over 4{/tex}
Number of birds coupled along = {tex}{x^2\over 9} +{ x^2\over 4}{/tex}
Number of birds moves on hill ={tex}7\sqrt {x^2} = 7x {/tex}
Number of birds remaining on trees = 56 Now,
{tex}{x^2\over 4} + {x^2\over 4} +{x^2\over 9} + 7x +56 = x^2{/tex}
=> {tex}=> {11x^2\over 18} + 7x + 56 = x^2{/tex}
{tex}=> 7x + 56 = {7x^2\over 18}{/tex}
{tex}=> x + 8 ={x^2\over 18}{/tex}
=> x^{2 }18x 144 = 0
=> x^{2}  24x +6x 144 = 0
=> x(x24) +6(x24) = 0
=> (x24)(x+6) = 0
x = 24 or 6
so Number of birds = 24^{2 }= 576
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A man stands at a point X on the bank XY of a river with straight and parallel bank, and observes that the line joining X to a points Z on the opposite Bank makes an angle of 30° with XY. He then goes along the bank a distance of 200 metre to Y and find that the angle ZYX 60°. Find the breadth of the river.

Answers:

Let the man 200 m to Y till point P.
Then, tan 30^{o} = ZY / XY
1/√3 = ZY / (200 + PY)
ZY = (200 + PY)/√3. .........(i)
Again, tan 60^{o} = ZY/PY
ZY = √3 PY .......(ii)
Equatind eq.(i) and (ii), we get,
(200 + PY)/√3 = √3 PY
200 + PY = 3PY
2PY = 200
PY = 100m
Therefore width of the river = XY = 200+ 100 = 300 m
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The perimeter of the ends of the frustum of the cone are 207.24 cm and 169.56 cm .if the height of the frustum be 8 cm,find the whole surface area of the frustum .

Answers:

At last, after multiplication the answer is not 7392.52.
The correct answer is 7592.52 sq.cm
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Ans. Let R and r be the radii of two ends of frustum.
Then
{tex}=> 2\times 3.14 \times R = 207.24{/tex}
=> R = 33 cm
{tex}=> 2\times 3.14 \times r = 169.56{/tex}
=> r = 27 cm
Height of frustum h = 8cm
Slant Height of Frustum l = {tex}\sqrt {(Rr)^2 + h^2}{/tex}
= {tex}\sqrt {(3327)^2+(8)^2} = \sqrt {36+64}{/tex}
=> 10
TSA of frustum = {tex}\pi l(R+r) +\pi R^2 +\pi r^2{/tex}
=> {tex}\pi [l(R+r) + R^2+r^2]{/tex}
=> {tex}3.14[10(33+27) + 1089 + 729]{/tex}
=> 3.14[600+1089+729]
=> 3.14 (2418)
=> 7392.52cm^{2}
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Water is flowing at a rate of 0.7 m/second through circular pipe whose internal diameter is 2 cm into a cylindrical tank,the radius of whose base is 40 cm.determine the increase in the level of water in half an hour

Answers:

Ans. Let water Level Rise in Tank = x cm
Radius of tank =
Speed of water = 0.7m/s = 70cm/s
Diameter of Pipe = 2 cm
Radius of Pipe = 1cm
Volume of water flowing In 1 Sec = {tex}{22\over 7}\times 1\times 1\times 70 = 220cm^3{/tex}
Volume of water flowing in half an hour =
{tex}220\times 1800 = 396000cm^3{/tex}
This Volume = Volume Of Tank
{tex}=> 396000 = {22\over 7}\times 40\times 40\times x{/tex}
=> x = 78.75 cm
Water Level increase in tank = 78.75cm
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Four equal circles are described at four corners a square so that each touches two of the others . The shaded area enclosed between the circles is 24/7 Cm square. Find the radius of each circle

Answers:

Ans.
Let ABCD be the square. Let r be the radius of each of the 4 equal circle that are described at each vertices of square ABCD.
Now, AB = BC = CD = DA = 2r
We know that each angle of a square is a right angle.
{tex}=> \angle A= \angle B= \angle C= \angle D = 90{/tex}
Now, area of a quadrant of the circle having centre at A = {tex}{90\over 360}\pi r^2 = {1\over 4}\pi r^2{/tex}
area of 4 quadrants = {tex}4\times {1\over 4}\pi r^2 = \pi r^2{/tex}
Area of Square = {tex}(2r)^2 = 4r^2{/tex}
Area of Shaded Region = Area of Square  Area of 4 quardants
{tex}=> {24\over 7} = 4r^2  \pi r^2 = r^2(4{22\over 7}){/tex}
{tex}=> {24\over 7} = r^2\times {6\over 7}{/tex}
{tex}=> r^2 = 4 => r = 2cm{/tex}
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A right triangle having sides 15 cm and 20 cm is made to revolve about it's hypotenuse .find the volume and surface area of the double cone so formed .

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Ans.
Consider the following right angled triangle ABC is rotated through its hypotenuse AC
BD {tex}\perp{/tex} AC. In this case BD is the radius of the double cone generated.
Using Pythagoras theorem for {tex}\triangle ABC{/tex}
=> AC^{2} = AB^{2} + BC
=> AC^{2} = 225 + 400 = 625
=> AC = 25 cm
Let AD = x cm CD = (25 – x) cm
Using Pythagoras theorem in {tex}\triangle ABD{/tex}
=> AD^{2} +BD^{2} = AB^{2}
=> x^{2} + BD^{2}= 225
=> BD^{2}= 225 – x^{2} …………(1)
Using Pythagoras theorem for {tex}\triangle CBD{/tex}
=> BD^{2 }+ CD^{2 }= BC^{2}
=> BD^{2} +(25x)^{2}= 400
=> BD^{2} = 400  (25x)^{2} …………(2)
From (1) & (2)
=> 225  x^{2}= 400  625  x^{2 }+ 50x
=> 50x = 450
=> x = 9 cm
AD = 9, CD = 16
From (1) BD = 12
Surface area of the double cone formed = L.S.A of upper cone + L.S.A of the lower cone
{tex} => \pi \times BD \times AB + \pi \times BD \times BC{/tex}
{tex} => \pi \times BD \times (AB + BC){/tex}
{tex}=> \pi \times 12(20+15){/tex}
{tex}=> \pi \times 12\times 35{/tex}
{tex}=>420 \pi {/tex}cm^{2}
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In an ap , the sum of first n terms is 3n^{2}/2 + 13n/2 . Find the 25^{th } term.

Thank u

Please send answer for this question.
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Answers:

S_{n} = 3n^{2}/2 + 13n/2
S_{1} = a = 3/2 + 13/2 = 8
S_{2} = a + a_{2} = 12/2 + 26/2 = 19
a_{2} = 19  8 = 11
Then,. d = 11 8 = 3
Now, a_{25} = a +(25  1)d
= 8 + 24 × 3
= 80
The 25th term is 80
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P and Q are centres of circles of radii 9 cm and 2 cm respectively . PQ is 17 cm . R is the centre of the circles of radius X cm which touches the above circle externally . Given that angle PRQ is 90 degree .write an equation in X and solve it.

Please send solution for this question. I have maths exam tommorow
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Answers:

Ans. If two circles touch each other externally, then sum of their radii is equal to the distance between their centres.
RP = x+9
RQ= x+2
PQ = 17
Given, angle PRQ= 90, So In Triangle PRQ => RP^{2} + RQ^{2} = PQ^{2 }…(using Pythagoras)
=> (x+9)^{2} + (x+2)^{2}= (17)^{2}
=> x^{2} + 18x + 81 + x^{2 }+ 4x + 4 = 289
=> 2x^{2 }+ 22x 204 =0
=> x^{2} + 11x 102 = 0
On Solving We get
x = 17 and 6
So radius of circle is 6cm.
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ABC is an isosceles triangle with AB=AC. D is the midpoint of AC. Taking BD as diameter a circle is drawn which meets AB at E. Prove that AE=1/4AC

Answers:

Ans.
Given: ABC is isosceles triangle. AB = AC. D is midpoint of AC.
To prove : {tex}AE = {1\over 4}AC{/tex}
Proof:
AB = AC …………(1) (Given)
AD = AC/2 ……………(3) (d is midpoint)
As AD is tangent to the triangle and AEB is secant then,
{tex}=> AE \times AB = AD^2{/tex}
{tex}=> AE \times AC = ({AC\over 2})^2{/tex}
{tex}=> AE \times AC = {AC^2\over 4}{/tex}
{tex}=> AE = {AC\over 4}{/tex}
Hence Proved.
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In a family of 3children the probability of having at least one boy is :

<hr/>
Aarohi mehra, Can u please show me the method of how u have solved? ¿
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Answers:

Ans. Three children can be as follow:
BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG.
No of total cases = 8
No of favorable cases = 7
Probability = 7/8
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 7/8
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