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  • If  and <img alt="\mathrm{\tan (A-B) = \frac{1}{\sqrt3}, \ 0\degree < A+B < 90\degree, A

If \mathrm{\tan (A+B) = 1} and \mathrm{\tan (A-B) = \frac{1}{\sqrt3}, \ 0\degree < A+B < 90\degree, A >B} then find the values of A and B.

 

 
 
 
 
 

Answers (1)

Given

        \tan (A +B) = 1

So,    A +B = 45\degree\qquad -(i)

        \tan (A -B) = \frac{1}{\sqrt3}

So,    A - B = 30\degree\qquad -(ii)

Solve (i) and (ii) we get    \angle A = 37\frac{1}{2}\degree \ or\ 37.5\degree

                                            \angle B = 7\frac{1}{2}\degree \ or\ 7.5\degree

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Safeer PP

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