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Sandhya Sodhani 6 years, 11 months ago
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Sia ? 6 years, 6 months ago
Here we are given that
{tex}\therefore \quad A B + A D = B C + C D{/tex}
or, {tex}AD = BC+CD-AB{/tex}
or, {tex}AD =h+d-x{/tex}
In the right angled {tex}\Delta A C D,{/tex}
{tex}A D ^ { 2 } = A C ^ { 2 } + D C ^ { 2 }{/tex}
or, {tex}( h + d - x ) ^ { 2 } = ( x + h ) ^ { 2 } + d ^ { 2 }{/tex}
or, {tex}( h + d - x ) ^ { 2 } - ( x + h ) ^ { 2 } = d ^ { 2 }{/tex}
{tex}( h + d - x - x - h ) ( h + d - x + x + h ) = d ^ { 2 }{/tex}
{tex} Because \ a^2-b^2=(a-b)(a+b){/tex}
or, {tex}( d - 2 x ) ( 2 h + d ) = d ^ { 2 }{/tex}
or, {tex}2 h d + d ^ { 2 } - 4 h x - 2 x d = d ^ { 2 }{/tex}
or, {tex}2hd = 4hx+2xd{/tex}
{tex}2hd= 2x(2h+d){/tex}
Hence {tex}x = \frac { h d } { 2 h + d }{/tex}
Posted by Yogesh Parihar 6 years, 11 months ago
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Posted by Peri Nabam 6 years, 11 months ago
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Sreenandu Hari 6 years, 11 months ago
Posted by Yuvraj Mandavi 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
We have,
x2 + y2 + z2 = r2sin2{tex}\alpha{/tex}cos2{tex}\beta{/tex} + r2sin2{tex}\alpha{/tex} sin2{tex}\beta{/tex} + r2cos2{tex}\alpha{/tex}
= r2sin2{tex}\alpha{/tex} (cos2{tex}\beta{/tex} + sin2{tex}\beta{/tex}) + r2cos2{tex}\alpha{/tex}
= r2sin2{tex}\alpha{/tex} + r2cos2{tex}\alpha{/tex} [{tex}\because{/tex} cos2{tex}\beta{/tex} + sin2{tex}\beta{/tex} = 1]
= r2(sin2{tex}\alpha{/tex} + cos2{tex}\alpha{/tex}) = r2 [{tex}\because{/tex} sin2{tex}\alpha{/tex} + cos2{tex}\alpha{/tex} = 1].
Hence, (x2 + y2 + z2) = r2.
Posted by Akash Kumar 6 years, 11 months ago
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Neha Narula 6 years, 11 months ago
Neha Narula 6 years, 11 months ago
D.J Alok 6 years, 11 months ago
Posted by Rani Kaurr 6 years, 11 months ago
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Gaurav Seth 6 years, 11 months ago
Here you can see 10, 13 , 16 , .....94 are in AP where first term , a = 10 and common difference , d = 94 .
From AP nth term formula ,
Tn = a + (n -1)d
94 = 10 + (n -1)3
84/3 = n -1 ⇒n = 29
Hence, there are 29 terms between 10 and 94
It means middle term is (n + 1)/2 th term
e.g., middle term = (29+1)/2 = 15th term
So, T₁₅ = a + (15 - 1)d = 10 + 14 ×3 = 52
Hence, 15th term is 52
here middle is 15th , it means 14 terms are in left side of it and 14 terms are in right side of it.
So, sum of first 14 term {left side } = 14/2 [ 2 × 10 + (14 -1)×3 ] [∵Sn = n/2[2a + (n-1)d]
= 7[20 + 39] = 7 × 59 = 413
Now, sum of last 14 term { right side of middle term } = S₂₉ -[ S₁₄ + T₁₅ ]
= 29/2[2 × 10 + (29-1) × 3 ] - 413 - 52
= 29/2[20 + 84 ] - 465
= 1508 - 465
= 1043
Posted by Khushboo . 6 years, 11 months ago
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Gaurav Seth 6 years, 11 months ago
To prove :
sin⁸x/a³ + cos⁸x/b³ =?
Solution:
Sin⁴x/a + cos⁴x/b = 1/a+b
As we know that Sin²x +Cos²x=1
∴Cos²x=1-sin²x
Cos⁴x=(Cos²x)²=(1-sin²x)²=1+sin⁴x-2sin²x
⇒Sin⁴x/a +[1+sin⁴x-2sin²x]/b =1/a+b
⇒Sin⁴x/a+1/b + sin⁴x/b -2sin²x/b=(1/a+b)
⇒Sin⁴x/a+ sin⁴x/b -2sin²x/b=(1/a+b)-1/b
⇒Sin⁴x(1/a +1/b)-2sin²xa/ab=[b-a-b]/b(a+b)
⇒Sin⁴x(1/a +1/b)-2asin²x/ab=-a/b(a+b)
⇒Sin⁴x(a+b)/ab-2asin²x/ab=-a/b(a+b)
⇒(a+b)²(Sin²x)² - 2a(a+b)sin²x= - a²
⇒(a+b)²(Sin²x)² - 2a(a+b)sin²x + a² =0
This equation is similar to a²-2ab+b²=0 ⇒(a-b)²
[(a+b) Sin²x-a]²=0
sin²x=a/a+b
Cos²x=1-sin²x= 1- a/(a+b)= b/a+b
∴sin²x=a/a+b and Cos²x= b/a+b
Now :
sin⁸x=a⁴/(a+b)⁴ and Cos⁸x= b⁴/(a+b)⁴
sin⁸x/a³= a/(a+b)⁴ and Cos⁸x/b³=b/(a+b)⁴
Now
sin⁸x/a³ + Cos⁸x/b³ = a/(a+b)⁴ + b/(a+b)⁴
=a+b/(a+b)⁴
⇒sin⁸x/a³ + Cos⁸x/b³=1/(a+b)³
Posted by Dolma Lhamo 6 years, 11 months ago
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Gaurav Seth 6 years, 11 months ago
angle A + angle C = 180 deg. and also angle B + angle D = 180 deg.
Now 2x + 4 + 2y + 10 = 180 => x+y = 83 deg.
similarly, y+3+4x-5 = 180 => 4x+y = 182 deg.
solving the two simultaneous equations in x and y we get,
3x = 99 deg; x= 33 deg
y = 50 deg.
now calculate A,B,C and D.
A = 70 deg => C = 110 deg
B = 53 deg => D = 127 deg
. . 6 years, 11 months ago
Posted by Priyanshu Pandey 6 years, 11 months ago
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Posted by Deepika Sahu 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
The given equation can be rewritten as
{tex}\frac { a ( x - a ) + b ( x - b ) } { ( x - b ) ( x - a ) } = 2{/tex}
a(x -a) + b(x- b) = 2[x2 - (a + b)x + ab]
ax- a2 + bx - b2 = 2x2- 2(a + b)x + 2ab
2x2- 3(a + b)x + (a + b)2 = 0
2x2 - 2(a + b) x - (a + b)x + (a + b)2 = 0
{tex}2x[x-(a+b)]-(a+b)[x-(a+b)]{/tex}
[2x - (a + b)] [x - (a + b)] = 0
{tex}x = a + b , \frac { a + b } { 2 }{/tex}
Posted by Kavya Mecharla 6 years, 11 months ago
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Shashikant Yadav 6 years, 11 months ago
Posted by Kanchan Kanchan 6 years, 11 months ago
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Deepanshu Singh 6 years, 11 months ago
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