sin(A+B)=√3/2=cos(A-B). Find A and B.
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hence value of A=45 ,and value of B=15
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Hello users ...
We know that:
sin 60° = √3 / 2
And
cos 30° = √3 / 2
Solution:-
here,
Given that;
sin (A + B) = √3 / 2
=> sin(A + B) = sin 60°
=> A + B = 60° ......... (1.)
And
cos (A-B) =√3 / 2
=> cos ( A - B) = cos 30°
⇒ A-B = 30°.......(2)
Now, solving ....
by adding equation (1.) and (2.)
we get
(A + B) + (A - B) = 60° + 30°
=> 2A = 90°
=> A = 90° /2 = 45°
By subtracting (2.) from (1.)
we get
(A + B) - ( A - B) = 60° - 30°
=> A + B - A + B = 30°
=> 2B = 30°
=> B = 30° / 2 = 15°
Hence;
∠A = 45°
And
∠B = 15° Answer
Hope it helps :)
We know that:
sin 60° = √3 / 2
And
cos 30° = √3 / 2
Solution:-
here,
Given that;
sin (A + B) = √3 / 2
=> sin(A + B) = sin 60°
=> A + B = 60° ......... (1.)
And
cos (A-B) =√3 / 2
=> cos ( A - B) = cos 30°
⇒ A-B = 30°.......(2)
Now, solving ....
by adding equation (1.) and (2.)
we get
(A + B) + (A - B) = 60° + 30°
=> 2A = 90°
=> A = 90° /2 = 45°
By subtracting (2.) from (1.)
we get
(A + B) - ( A - B) = 60° - 30°
=> A + B - A + B = 30°
=> 2B = 30°
=> B = 30° / 2 = 15°
Hence;
∠A = 45°
And
∠B = 15° Answer
Hope it helps :)
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