If tan (A + B) = √3 and tan (A – B) = 1

If tan (A + B) = √3 and tan (A – B) = 1 and A, B (B < A) are acute angles, find the values of A and B.

Answer :

tan (A + B) = √3

So, tan (A + B) = tan 60° [Since, tan 60° = √3]

⇒ A + B = 60° …(i)

tan (A – B) = 1

tan (A – B) = tan 45° [tan 45° = 1]

⇒ A – B = 45° …(ii)

From equation (1) and (2), we get

A + B + A – B = 60° + 45°

⇒ 2A = 105o

⇒ A = 52½o

on substituting the value of A in equation (i), we get

52½o + B = 60°

⇒ B = 60° – 52½o = 7½o

Therefore, the value of A = 52½and B = 7½o

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