Question

cos(A-B) =√3/2sin(A+B)=√3/2.where(a+b)&(a-b) are acute angle.Find a-b

Answer 1

we know that cos30=root3/2

$so \:because \: cos(a - b) = \frac{ \sqrt{3} }{2} \\ so \: cos(a - b) = cos30 \\ so \: (a - b) = 30$

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Answer 2

Given,
cos(A-B)=√3/2, sin(A+B)=√3/2
So,
cos(A-B) = √3/2
cos(A-B) = cos 30 [ cos 30 = √3/2]
A-B = 30 ... (1)
Similarly
sin(A+B) = √3/2
sin(A+B) = sin 60 [ sin 60 = √3/2]
A+B = 60 ...(2)
adding eq. (1) and (2) we get,
A+B = 60
A-B = 30
_______
2A = 90
OR A = 45
putting value of A in eq. (2) we get,
45+B = 60
B = 60-45
B = 15
So, value of A-B will be 45-15 i.e. 30.
Ans. Value of A-B is 30.