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Yogita Ingle 6 years, 3 months ago
Given, cos 45/(sec 30 + cosec 30)
= (1/√2)/(2/√3 + 2) {since cos 45 = 1/√2, sec 30 = 2/√3, cosec 30 = 2}
= (1/√2)/{(2 + 2√3)/√3}
= (√3/√2)/(2 + 2√3)
= √3/{√2*(2 + 2√3)}
= √3/(2√2 + 2√3*√2)
= √3/(2√2 + 2√6)
So, cos 45/(sec 30 + cosec 30) = √3/(2√2 + 2√6)
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Sia ? 6 years, 3 months ago
- By Elimination method,
The given system of equations is :
3 x - 5 y - 4 = 0............(1)
9 x = 2 y + 7
9 x - 2 y - 7 = 0.............(2)
Multiplying equation (1) by 3, we get
9 x - 15 y - 12 = 0.............(3)
Subtracting equation (3) from equation (2) , we get
13 y + 5 = 0
{tex}\Rightarrow \quad 13 y = - 5 \Rightarrow y = \frac { - 5 } { 13 }{/tex}
Substituting this value of y in equation (1), we get
{tex}3 x - 5 \left( \frac { - 5 } { 13 } \right) - 4 = 0{/tex}
{tex}\Rightarrow \quad 3 x + \frac { 25 } { 13 } - 4 = 0 \Rightarrow 3 x - \frac { 27 } { 13 } = 0{/tex}
{tex}\Rightarrow \quad 3 x = \frac { 27 } { 13 } \Rightarrow x = \frac { 9 } { 13 }{/tex}
So, the solution of the given system of equation is
{tex}x = \frac { 9 } { 13 } , y = \frac { - 5 } { 13 }{/tex} - By Substitution method:
The given system of equation is:
3 x - 5 y - 4 = 0.............(1)
9 x = 2 y + 7...................(2)
From equation (2),
{tex}x = \frac { 2 y + 7 } { 9 }{/tex}..................(3)
Substituting this value of x in equation(1), we get
{tex}3 \left( \frac { 2 y + 7 } { 9 } \right) - 5 y - 4 = 0{/tex}
{tex}\Rightarrow \quad \frac { 2 y + 7 } { 3 } - 5 y - 4 = 0{/tex}
{tex}\Rightarrow \quad 2 y + 7 - 15 y - 12 = 0{/tex}
{tex}\Rightarrow{/tex} -13y - 5 = 0
{tex}\Rightarrow{/tex} 13y = -5
{tex}\Rightarrow \quad y = \frac { - 5 } { 13 }{/tex}
Substituting this value of y in equation(3), we get
{tex}x = \frac { 2 \left( \frac { - 5 } { 13 } \right) + 7 } { 9 } = \frac { - \frac { 10 } { 13 } + 7 } { 9 } = \frac { - 10 + 91 } { 117 } = \frac { 81 } { 117 } = \frac { 9 } { 13 }{/tex}
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Sia ? 6 years, 3 months ago
Given cos A = cos B
Hence, {tex}\frac{{AC}}{{AB}} = \frac{{BC}}{{AB}}{/tex}
{tex} \Rightarrow AC = BC{/tex}
Since angle opposite to equal sides in a {tex}\Delta {/tex} are equal
{tex}\therefore \angle B = \angle A{/tex}
Hence proved
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Sia ? 6 years, 3 months ago
Suppose x litres of 50% solution be mixed with y litres of 25% solution.
According to the question,
50% of x + 25% of y = 40% of 10
{tex}\Rightarrow \frac { 50 } { 100 } x + \frac { 25 } { 100 } y = \frac { 40 } { 100 } ( 10 ){/tex}
{tex}\Rightarrow{/tex} {tex}50x + 25y = 40(10){/tex}
{tex}\Rightarrow{/tex} {tex}2x + y = 16{/tex}..........(i)
The amount of each solution adds to {tex}10\ litres{/tex},
{tex}x + y = 10{/tex}.......(ii)
Subtract (ii) from (i),
{tex}\Rightarrow{/tex} {tex}x = 6{/tex}
Substituting {tex}x = 6{/tex} in (i), we get {tex}y = 4{/tex}
{tex}\therefore{/tex}{tex}6\ litres{/tex} of 50% solution is to be mixed with {tex}4\ litres{/tex} of 25% solution
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