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Asked by krishpatel20108 | 01 Jan, 2023, 07:56: PM
254 x + 309 y = -55   ...................................... (1)

309 x + 254 y = 55  .........................................(2)

By adding eqn.(1) and (2) , we get 563 x + 563 y = 0

Hence x + y = 0 or   x = -y

if we substitute x = -y in eqn.(1)  , we get y = -1, hence x = +1

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Question 23

(x2 - 3 ) = 0

( x + √3 ) ( x - √3 ) = 0

hence x = ± √3

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Question 24

If all the points are collinear , then a straight line ax +by + c = 0 is paasing through all the points

a x + b y + c = 0

Let us divide the above equation by a, then we get

x + (b/a ) y + ( c /a ) = 0  ..............................(1)

let ( b/a ) = p and ( c/a ) = q

Then using above substitution , we get

x + py = -q  ................................. (2)

if the point (1, 2 ) is on the line , then  1 + 2p = -q

2p +q = -1  ........................................ (3)

if the point (7, 0 ) is on the line , then  7 = -q

By substituting q = -7 in eqn.(3) , we get p = 3

Hence the relation between x and y is obtained from eqn.(2)

x + 3 y = 7

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Question 25

Fig.(1) shows two concentric eircles. Radius of inner circle is OA = b and
radius of outer circle OB = a.

Cord CB of bigger circle is touching smaller circle at A , Hence CB is tangent

CB = 2 AB =
-------------------------------------------------------

Fig.(2) shows that a tangent is drawn from P to a circle and this tangent meets the circle at Q.

QR is diamaeter of circle.

POR = 110o ( given )
POQ = 180 - POR = 70o
ΔPOQ is right angled triangle  and Q is right angle .
Hence OPQ = 20o
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Question 26

ΔABC is right angled triangle right angled at C.

AB = 29 units ( given )

BC = 21 units ( given )

AC =
cosθ = BC / AB = 21/29

sinθ = AC/AB = 20/29

cos2θ + sin2θ = (21/29)2 + (20/29)2 = 1

cos2θ - sin2θ = (21/29)2 - (20/29)2 = 0.048

Answered by Thiyagarajan K | 01 Jan, 2023, 11:37: PM

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