If $ \tan (A-B)=\frac{1}{\sqrt{3}} $ and $ \tan (A+B)=\sqrt{3}, 0^{\circ} < A+B \leq 90^{\circ}, A>B $ find $ A $ and $ B $.

Given:

\( \tan (A-B)=\frac{1}{\sqrt{3}} \) and \( \tan (A+B)=\sqrt{3}, 0^{\circ} < A+B \leq 90^{\circ}, A>B \)

To do:

We have to find $A$ and $B$.

Solution:  

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

$\tan (A-B)=\tan 30^{\circ}$          (Since $\tan 30^{\circ}=\frac{1}{\sqrt3}$)       

$\Rightarrow  A-B=30^{\circ}$......(i)

$\tan (A+B)=\sqrt3$

$\tan (A+B)=\tan 60^{\circ}$             (Since $\tan 60^{\circ}=\sqrt3$)

$\Rightarrow A+B=60^{\circ}$

$\Rightarrow  A=60^{\circ}-B$........(ii)

Substituting (ii) in (i), we get,

$60^{\circ}-B-B=30^{\circ}$

$\Rightarrow  2B=30^{\circ}$

$\Rightarrow  B=\frac{30^{\circ}}{2}$

$\Rightarrow  B=15^{\circ}$

$\Rightarrow  A=60^{\circ}-15^{\circ}=45^{\circ}$

The values of $A$ and $B$ are $45^{\circ}$ and $15^{\circ}$ respectively. 

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Updated on: 10-Oct-2022

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