Question

if cos(A+B)=2sin(A-B)=1 ,find the value of acute angle A and B

Answer 1

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


$2 \times 1 \div 2 = 1$
therefore

A=60

B=30

are the acute angles

Answer 2

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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