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CBSE Class 10 Answered

maths
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Asked by krishpatel20108 | 01 Jan, 2023, 07:56: PM
answered-by-expert Expert Answer
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
 
begin mathsize 14px style A B space equals space square root of left parenthesis O B right parenthesis squared minus left parenthesis O A right parenthesis squared end root space equals space square root of a squared minus b squared end root end style
CB = 2 AB = begin mathsize 14px style 2 space square root of a squared minus b squared end root end style
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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.
 
begin mathsize 14px style angle end stylePOR = 110o ( given )
begin mathsize 14px style angle end stylePOQ = 180 - begin mathsize 14px style angle end stylePOR = 70o
ΔPOQ is right angled triangle  and begin mathsize 14px style angle end styleQ is right angle .
Hence begin mathsize 14px style angle end styleOPQ = 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 = begin mathsize 14px style square root of A B squared space minus space B C squared end root space equals space square root of left parenthesis 29 right parenthesis squared minus left parenthesis 21 right parenthesis squared end root space equals space 20 space u n i t s end style
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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