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Let two tangent PT and QT are drawn to circle of centre O as shown in figure. Both the given tangents PT and QT touch to the circle at P and Q respectively.

We have to proof : length of PT = length of QT Construction :- draw a line segment ,from centre O to external point T { touching point of two tangents } .

Now ∆POT and ∆QOT We know, tangent makes right angle with radius of circle. Here, PO and QO are radii . So, ∠OPT = ∠OQT = 90° Now, it is clear that both the triangles ∆POT and QOT are right angled triangle.nd a common hypotenuse OT of these [ as shown in figure ]

Now, come to the concept , ∆POT and ∆QOT ∠OPT = OQT = 90° Common hypotenuse OT And OP = OQ [ OP and OQ are radii]So, R - H - S rule of similarity ∆POT ~ ∆QOT Hence, OP/OQ = PT/QT = OT/OT PT/QT = 1 PT = QT [ hence proved]

S I= p*t*r/100p= SI*100/t*rp = 2080*100/3.25*16p= 208000/52p= 4000

P=620n=4 monthsr=8pnr÷100=198.4interest=198.4amount=198.4+620=818.4

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i) put n = 1,2,3 first term is 2*(1)-3 = -1 2nd term is 2*(2)-3 = 1 3rd term is. 2*(3)-3 = 3

here Common difference is same for all the terms that is 1-(-1) = 2. so , it is an A.P

ii) similarly for n²+1 first term is 2 second term is (2)²+1 = 53rd term is (3)²+1 = 10so, here Common difference is not same..5-1 = 4 but 10-5 = 5 , so it is not an A.P

1)let one no.be xso another no x+4according to conditionx+(x+4)=482x+4=482x=44x=22

therefore one no.is 22 and another no.is (22+4)=26

2)let the nos be 8x and 9xaccording to conditionif they differ by 78x-9x=7-x=7x=-7

so,the nos are8x=8*-7=-569x=9*-7=-63

3)

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thnku so much dear

thanks for the answer

8x+4=60is the equation of above data.

8x+4=60;X=7

Thnx for answering

two equations can be

x+3 = 1 2x+7= 3

thanks

the graphical method is to solve question by graph

hey friend I don't know the answer and please like my answer

squaring both sides,

... a² cos² x + b² sin² x - 2ab sin x cos x = c²

∴ a² ( 1 - sin² x ) + b² ( 1 - cos² x ) - 2ab sin x cos x = c²

∴ a² - a² sin² x + b² - b² cos² x - 2ab sin x cos x = c²

∴ a² + b² - c² = a² sin² x + b² cos² x + 2ab sin x cos x

∴ a² + b² - c² = ( a sin x + b cos x )²

∴ a sin x + b cos x = ± √( a² + b² - c² ). .................................. Q.E.D....

Find volume of metallic pipe take radius 7cm then again find out volume with inner radius (6cm)Now subtract the volume (outer - inner)And multiply with density of metal

Volume of the metallic cylindrical pipe = 748 Cm³

length of the cylindrical pipe h =14 cm

external radius of the pipe R = 9 cm

let the internal radius of the cylindrical pipe be r cm.

therefore the volume of the metallic pipe= π(R1²-r2²)h

therefore

thickness of the metallic pipe = R-r=9-8=1 cm

-7`3

5x * 6x = 4320

30x² = 4320

x² = 144

x = 12

Second and third = 5*12 = 60 and 6*12 = 72

Ist and 2nd numbers = 3*a : 4*a

3*a : 72

3*a : 4*18

3*18= 54 and 72

Difference =72-12=60

226 a3 b 3Hope it will help you

(8a+7b)(8a+7b); is the correct answer of the given

The trigonometric ratio is tan.

Let unkown number be x

x + 12 = 33

x = 33 - 12

x = 21

a)x -5 = 05x - 25 = 0x + 10 = 15

b)3y = 13y + 4 = 53y + 7 = 8

c)y - 15 = 03y = 45y + 10 = 25

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1) VIII < IX2) XXI > XIX

But, how did u get dis divisor??

1

this is not my question

root over (2+√x) = 32+√x = 9√x = 7x = 7² = 49option (D) is correct