Question : If $\tan(A+B)=\sqrt{3}$, and $\tan(A-B)=\frac{1}{\sqrt{3}}$, $\angle A+\angle B<90°$; $A\geq B$ then $\angle A$ is:
Option 1: 90°
Option 2: 30°
Option 3: 45°
Option 4: 60°
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Correct Answer: 45°
Solution :
Given: $\tan(A+B)=\sqrt{3}$
⇒ $A+B=60°$ ------(1)
$\tan(A-B)=\frac{1}{\sqrt{3}}$
⇒ $A-B=30°$ ------(2)
Adding equations (1) and (2), we have,
$2A=90°$
$\therefore A=45°$
Hence, the correct answer is 45°.
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