Latest SELF Aptitude Question SOLUTION: Q. Solve 3SinA + 4CosA = 5 4CosA - 3SinA = ?

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Q. Solve
3SinA + 4CosA = 5
4CosA - 3SinA = ?

Akanksha Mishra (10 years ago)

Unsolved Read Solution (8)

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Akanksha Mishra  (10 years ago)

Thank you for ur try sorry ur solution is wrong i become able to solve it
i) 3cos(A) + 4sin(A) = 5 [Given]

ii) Let 4CosA - 3SinA = x, the value of which needs to be found out.

iii) Squaring the first one it is,
9cos²A + 16sin²A + 24cos(A)sin(A) = 25
4CosA - 3SinA = x
Similarly squaring the second one,
9sin²A + 16cos²A - 24cos(A)sin(A) = x²

Adding both, 9 + 16 = 25 + x²
[Since by identity, sin²A + cos²A = 1]

==> x² = 0
So x = 0
4CosA - 3SinA = 0

Is this solution Helpfull? Yes (3) | No (3)  

Akanksha Mishra  (10 years ago)

Thank you for ur try sorry ur solution is wrong i become able to solve it
i) 3cos(A) + 4sin(A) = 5 [Given]

ii) Let 3sin(A) + 4cos(A) = x, the value of which needs to be found out.

iii) Squaring the first one it is, 9cos²A + 16sin²A + 24cos(A)sin(A) = 25
Similarly squaring the second one, 9sin²A + 16cos²A - 24cos(A)sin(A) = x²

Adding both, 9 + 16 = 25 + x² [Since by identity, sin²A + cos²A = 1]

==> x² = 0
So x = 0

Thus 3sin(A) + 4cos(A) = 0

Is this solution Helpfull? Yes (2) | No (4)  

mohan lal garg  (10 years ago)

@akanksha mishra your solution valid for other question
u take
i) 3cos(A) + 4sin(A) = 5 [Given]
but in question 3SinA + 4CosA = 5
solution for question

3SinA + 4CosA = 5
satisfy
cosA=4/5,sinA=3/5
so
4CosA - 3SinA = =16/5-9/5=7/5
ans is 7/5

for your solution
3cos(A) + 4sin(A) = 5
cos a=3/5, sina 4/5
4CosA - 3SinA = 4*3/5-3*4/5=0
so please check your question

Is this solution Helpfull? Yes (1) | No   

RAKESH  (10 years ago)

3SinA + 4CosA = 5
or (3/5)sinA +(4/5)cosA = 1
or cos(53)sinA + sin(53)cosA = 1 [since, cos(53 degree)=3/5]
or sin(53+A)=1=sin(90)
or 53+A=90
or A=37
so 4CosA - 3SinA
= 4cos(37)-3sin(37)=4*(4/5)-3*(3/5)=7/5=1.4

Is this solution Helpfull? Yes | No (2)  

RAKESH  (10 years ago)

3SinA + 4CosA = 5
or (3/5)sinA +(4/5)cosA = 1
let (3/5)=sinB then 4/5=cosB
so eqn becomes
sinB*sinA + cosB*cosA =1
or cos(A-B)=1=cos(0 degree)
or A-B=0
or A=B
SO (3/5)=sinB=sinA & (4/5)=cosB=cosA
so 4CosA - 3SinA
=4*(4/5)-3*(3/5)=7/5
=1.4

Is this solution Helpfull? Yes | No (1)  

RAKESH  (10 years ago)

3SinA + 4CosA = 5
4CosA - 3SinA = x
adding both gives, 8cosA=(5+x) or cosA=(5+x)/8
subtraction gives, 6sinA=(5-x) or sinA=(5-x)/6
now squaring & adding these two
(cosA)^2+(sinA)^2=[(5+x)/8]^2+[(5-x)/6]^2
or 1=[(5+x)/8]^2+[(5-x)/6]^2
solving, x=1.4
4CosA - 3SinA =x=1.4

Is this solution Helpfull? Yes | No (1)  

RAKESH  (10 years ago)

let x=sinA & Y=sinB
then 3x+4y=5
also x^2+y^2=1
solving x=3/5,y=4/5
so 4CosA - 3SinA =4y-3x=4*(4/5)-3*(3/5)=7/5=1.4

Is this solution Helpfull? Yes | No (1)  

mugilu  (8 years ago)

3 is opposite side of triangle and 4 is base and 5 is hypotenuse
of right angled of triangle let be think.......
Then....4cosA-3cosa= 7. =fix 1.2 answer
-------
5

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