if sin (A+B) =1 and cos(A-B)=√3/2, where A>B, then B =?
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sin(A+B) = 1
=> sin(A+B) = sin 90
=> A+B = 90 ____ equ. no. 1
cos(A-B) = √3/2
=> cos(A-B) = cos 30
=> A-B = 30 ____ equ. no. 2
Subtracting equ. no. 1 & 2
(A+B) - (A-B) = 90 - 30
=> A+B-A+B = 60
=> 2B = 60
=> B = 30 [Ans.]
Hope it will help you .
=> sin(A+B) = sin 90
=> A+B = 90 ____ equ. no. 1
cos(A-B) = √3/2
=> cos(A-B) = cos 30
=> A-B = 30 ____ equ. no. 2
Subtracting equ. no. 1 & 2
(A+B) - (A-B) = 90 - 30
=> A+B-A+B = 60
=> 2B = 60
=> B = 30 [Ans.]
Hope it will help you .
lonely3:
but correct answer is 60
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Answered by
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Hi ,
Sin ( A + B ) = 1
Sin(A + B ) = sin 90°
A + B = 90 ---( 1 )
Cos ( A - B ) = √3/2
Cos( A - B ) = cos 30°
A - B = 30° ---( 2 )
Add equations ( 1 ) & ( 2 ) , we get
2A = 120°
A = 60°
Substitute A = 60° in equation ( 1 ) ,
60° + B = 90°
B = 30°
Therefore ,
A = 60°
B = 30°
I hope this helps you.
: )
Sin ( A + B ) = 1
Sin(A + B ) = sin 90°
A + B = 90 ---( 1 )
Cos ( A - B ) = √3/2
Cos( A - B ) = cos 30°
A - B = 30° ---( 2 )
Add equations ( 1 ) & ( 2 ) , we get
2A = 120°
A = 60°
Substitute A = 60° in equation ( 1 ) ,
60° + B = 90°
B = 30°
Therefore ,
A = 60°
B = 30°
I hope this helps you.
: )
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