Question

if tan(A-B) =1/√3 and tan (A+B) =1 find A​

Answer 1

Answer:

8.2, 3 If tan (A + B) = √3 and tan (A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B , find A and B. Given that Our equations are A + B ...

Answer 2

Step-by-step explanation:

$\tan(a - b) = \frac{1}{ \sqrt{3} }$

$\tan(a + b) = 1$

Add both equations, we get

$\tan(a - b) + \tan(a + b) = \frac{1}{ \sqrt{3} } + 1$

$\tan(a) + \tan(a) + \tan(b) - \tan(b) = \frac{1 + \sqrt{3} }{ \sqrt{3} } .$

$2 \tan(a) = \frac{1 + \sqrt{3} }{ \sqrt{3} } .$

$\tan(a) = \frac{1}{2} \times \frac{1 + \sqrt{3} }{ \sqrt{3} } .$

$\tan(a) = 0.5 \times 1.58$

$\tan(a) = 0.79$

$a = \cot(0.79)$

$a = 38.3$

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