CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Given: XY and X'Y' at are two parallel tangents to the circle wth centre O and AB is the tangent at the point C,which intersects XY at A and X'Y' at B.
To prove: ∠AOB = 90º .
Construction: Join OC.
Proof:
In ΔOPA and ΔOCA
OP = OC (Radii )
AP = AC (Tangents from point A)
AO = AO (Common )
ΔOPA ≅ ΔOCA (By SSS criterion)
Therefore, ∠POA = ∠COA .... (1) (By C.P.C.T)
Similarly , ΔOQB ≅ ΔOCB
∠QOB = ∠COB .......(2)
POQ is a diameter of the circle.
Hence, it is a straight line.
Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180°
From equations (1) and (2), it can be observed that
2∠COA + 2∠COB = 180°
∴ ∠COA + ∠COB = 90°
∴ ∠AOB = 90°
Hence proved.

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Simplest form of 148/185.

Posted by Absolute Abi 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Abhay Modi 6 years, 10 months ago

2|148. and. 5|185 2|74. |37 |37 Therefore; 148/185 = (2×2×37)/(5×37) =2×2/5 = 4/5(answer)

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Solve 2x+3y=11 and 2x-4y=-24 and hence find the value of 'm' for which y =mx +3

Posted by Anjani Singh 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Parul Kakria 6 years, 10 months ago

For value of m y=mx+3 5=m(-2)+3 5-3=-2m 2=-2m 2/-2=m -1=m

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Parul Kakria 6 years, 10 months ago

2x+3y=11_I 2x-4y=-24_II Subrtacting II from I 2x+3y=11 2x-4y=-24 - + + --------------- 7y=35 y=35/7 y=5 Put y=5 in I 2x+3y=11 2x+3(5)=11 2x+15=11 2x =11-15 2x =-4 x =-4/2 x=-2 Therefore,x=-2 y=5

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The sum and the product of zeroes of a quadratic polynomial p(x) are 7 and 10 respective . Then find p(x)

Posted by Gaurika Singh 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Harshika Saxena 6 years, 10 months ago

P(x) = x^2 - 7x + 10

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Safeeq Ahamed 6 years, 10 months ago

Alpha + beta=7 Alpha *beta=10 P)(x) =k(x^2 - alpha+beta x+alpha *beta) =k(x^2 - 7x+10)

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Is there anyway to logout from this app if it is there then plz tell

Posted by Sw Kkk 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Sw Kkk 6 years, 10 months ago

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1+secQ-TanQ/1+secQ+tanQ

Posted by Zahraan Lateef 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

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Prove that : sinA- 2sin^A / 2cos ^ A - cos A = tan A

Posted by Anu Sharma 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

LHS = {tex}\frac{\sin A-2 \sin ^{3} A}{2 \cos ^{3} A-\cos A}{/tex}
= {tex}\frac{\sin A\left(1-2 \sin ^{2} A\right)}{\cos A\left(2 \cos ^{2} A-1\right)}{/tex}
= {tex}\frac{\sin A\left(1-2\left(1-\cos ^{2} A\right)\right)}{\cos A\left(2 \cos ^{2} A-1\right)}{/tex} [{tex}\because sin^2\theta=1-cos^2\theta{/tex}]

= {tex}\frac{\sin A\left(1-2+2cos ^{2} A\right)}{\cos A\left(2 \cos ^{2} A-1\right)}{/tex}

= {tex}\tan A \frac{\left(2 \cos ^{2} A-1\right)}{\left(2 \cos ^{2} A-1\right)}{/tex}
= tan A = RHS
Hence proved.

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If a sin 14 find other trigno rayio

Posted by Jashmon Baliar Singh 5 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

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The radius and slant height of a right circular cone are in the ratio 7:13 and its curved surface area is 286cm. Find its radius

Posted by Rajesh Kumar Didwadiya 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Aruneeraj K 6 years, 10 months ago

Radius is 7 cm Take radius and slant height as 7x and 13x So csa=3.14×r×l, 286=22/7*7x*13x, x²=1 x=1& radius =7

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So Idhar kuch mahol garam SA lag rha hein,,,, btw I am devangan sharma... Woh aise hi Nam rakh diya tha

Posted by Osama Bin Laden Zinda Hein 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

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हमारे जवान सिर्फ 2km तक अंदर गए और दहशत 200 km तक अंदर गयी …. …..jai hind ….

Posted by Heart ? Hacker 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Nitin Kumar?Hr35 Vala?(Narnaul)? 6 years, 10 months ago

Jai hind??????

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The angle of elevation of the top and the bottom of a 50m high building from the top of a tower are 45degree and 60degree respectively find the height of the tower and the horizantal distance between the tower and the building.(use√3=1.73)

Posted by Shareef Khan 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Harshika Saxena 6 years, 10 months ago

tan45=100/×+y ×+y=100 y=×+100 tan30=100/× 1/1.732=100/× ×=100×1.732 ×=173.2 Now , distance b/w them , y=×-100 y=173.2-100 y=73.2 , which is the required answer.......??

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0.68 (bar on 68)+0.73 (bar on 73)

Posted by Appu Enterprises 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

CM and RN are respectively the median of ∆ABC and∆PQR . If ∆ABC~∆PQR ,prove that i) ∆AMC~∆PNR

Posted by Aakash Kumar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Harshika Saxena 6 years, 10 months ago

As CM amd RN are medians ,so 2MA=AB AND 2PN=PQ. Now ,as ?️ABC〰️?️PQR then angleA = angleP and AB/PQ=AC/PR ,➡️2MA/2PN=AC/PR ➡️MA/PN= AC/PR , SO BY SAS Criteria of Similarity, ?️AMC〰️?️PNR....... HENCE PROVED ????

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7 years ago,varun's age was square of swati's age.3 yrs hence swati's age will be 2/5 of varun's age.find their present ages.

Posted by Nishita Gupta 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let seven years ago Swati's age be x years.
As per given condition
Seven years ago Varun's age was five times the square of Swati's age.
Then, seven years ago Varun's age was 5x² years.
Therefore, Swati's present age = (x + 7) years
And Varun's present age = (5x² + 7) years
Three years hence Swati's age = (x + 7 + 3) years = (x + 10) years
And three years hence Varun's age = (5x² + 7 + 3) years = (5x² + 10) years
As per given condition
Three years hence Swati's age will be {tex}\frac{2}{5}{/tex} of Varun's age.
Therefore, x + 10 = {tex}\frac{2}{5}{/tex}(5x² + 10)
{tex}\Rightarrow{/tex} x + 10 = 2(x² + 2)
{tex}\Rightarrow{/tex} x + 10 = 2x² + 4
{tex}\Rightarrow{/tex} 2x² - x - 6 = 0
{tex}\Rightarrow{/tex} 2x² - 4x + 3x - 6 = 0
{tex}\Rightarrow{/tex} 2 x ( x - 2 ) + 3 ( x - 2 ) = 0
{tex}\Rightarrow{/tex} (2x + 3) (x - 2) = 0
{tex}\Rightarrow{/tex} x - 2 = 0 [{tex}\because{/tex} 2x + 3 {tex}\neq{/tex} 0 as x > 0]
{tex}\Rightarrow{/tex} x = 2
Hence, Swati's present age = (2 + 7) years = 9 years
Varun's present age = (5 {tex}\times{/tex}2² + 7) years = 27 years