Question

if sin (3alpha-beta)=1 and cos(2alpha+beta)=1/2,then the values of alpha and beta will be respectively options are (a)60degree,30degree,(b)30degree,0degree,(c)60degree,0degree and (d)30degree,45degree

Answer 1

Answer:

$\sin((3 \alpha - \beta ) = 1 \\ sin(3 \alpha - \beta ) = \sin(90 ) \\ 3 \alpha - \beta = 90 \: \: \: \: \: eq \: 1 \\ \cos(2 \alpha + \beta ) = \frac{1}{2} \\ \cos(2 \alpha + \beta ) = \cos(60) \\ 2 \alpha + \beta = 60 \: \: \: eq2 \\adding \: eq \: 1 \: and \: 2 \\ so \\ 5 \alpha = 150 \\ \alpha = 30 \\ put \: \alpha = 30 \\ 2 \alpha + \beta = 60 \\ 2(30) + \beta = 60 \\ 60 + \beta = 60 \\ \beta = 0 \\ hope \: you \: understand \: \\ and \: follow \: me$

B is correct option

Answer 2

Question :

If sin(3A-B) = 1 and cos(2A+B) = 1/2 , then the values of A and B will be respectively :

a) 60°,30°

b) 30°,0°

c) 60°,0°

d) 30°,45°

Answer :

option (b)

30°,0°

Note :

sin0° = 0

sin30° = 1/2

sin45° = 1/√2

sin60° = √3/2

sin90° = 1

cos0° = 1

cos30° = √3/2

cos45° = 1/√2

cos60° = 1/2

cos90° = 0

Solution :

We have

=> sin(3A-B) = 1

=> sin(3A-B) = sin90°

=> 3A - B = 90°

=> B = 3A - 90° ------------(1)

Also,

=> cos(2A+B) = 1/2

=> cos(2A+B) = cos60°

=> 2A + B = 60°

=> B = 60° - 2A -----------(2)

Now,

From eq-(1) and eq-(2) , we have ;

=> 3A - 90° = 60° - 2A

=> 3A + 2A = 60° + 90°

=> 5A = 150°

=> A = 150°/5

=> A = 30°

Now,

Putting A = 30° in eq-(1) , we get ;

=> B = 3A - 90°

=> B = 3•(30°) - 90°

=> B = 90° - 90°

=> B = 0°

Hence,

Option (b) is correct , ie ; 30°,0°