if cos(A+B)=2sin(A-B)=1 ,find the value of acute angle A and B
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Answered by
0
cos(A+B)=2sin(A+B)=1
Take sin(A+B)=1
sin(60+30)=1
i.e,. sin 90=1
now take. 2sin(A-B) =1
2sin(60-30)=1
therefore
A=60
B=30
are the acute angles
Take sin(A+B)=1
sin(60+30)=1
i.e,. sin 90=1
now take. 2sin(A-B) =1
2sin(60-30)=1
therefore
A=60
B=30
are the acute angles
Answered by
4
cos ( A + B ) = 2 sin ( A - B ) = 1
Given : A and B are acute angles .
So : A + B = 90° ............................(1)
==> 2 sin ( A - B ) = 1
==> sin ( A - B ) = 1/2
==> sin ( A - B ) = sin 30° [ see the attachment table sin 30° = 1/2 ]
==> A - B = 30° ......................(2)
Adding 1 and 2 we get :
A + B + A - B = 90° + 30°
==> 2 A = 120°
==> A = 120°/2 = 60°
A - B = 30°
==> 60° - B = 30°
==> B = 60° - 30°
==> B= 30°
The values are :
A = 60°
B = 30°
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