If \(\tan (A + B) = \sqrt{3}\) and \(\tan (A - B) = \dfrac{1}{\sqrt{3}}\); \(0^\circ < A + B \leq 90^\circ\), \(A > B\), find \(A\) and \(B\).
\(\angle A = 45^\circ,\ \angle B = 15^\circ\)
Given: \(\tan(A+B)=\sqrt{3}\) and \(\tan(A-B)=\dfrac{1}{\sqrt{3}}\), with \(0^\circ<A+B\le 90^\circ\) and \(A>B\).
To find: \(A\) and \(B\).
Step 1: Convert each tangent value to its standard angle
We use the standard values:
\(\tan 60^\circ = \sqrt{3}\) and \(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\).
So we can write:
\(\tan(A+B)=\tan 60^\circ\)
\(\tan(A-B)=\tan 30^\circ\)
Step 2: Use the given range to fix the exact angles
Since \(0^\circ<A+B\le 90^\circ\), the angle \((A+B)\) must be an acute angle.
In the acute range, \(\tan\theta\) has only one standard match, so:
\(A+B = 60^\circ\).
Also, since \(A>B\), we have \(A-B>0\). The standard acute angle for \(\tan\theta = \dfrac{1}{\sqrt{3}}\) is \(30^\circ\), so:
\(A-B = 30^\circ\).
Student Note: Without the range information, \(\tan\theta\) could repeat every \(180^\circ\). The conditions given in the question help us choose the correct (acute) angles.
Step 3: Solve the two equations
We now have a pair of simple linear equations:
(1) \(A+B=60^\circ\)
(2) \(A-B=30^\circ\)
Add (1) and (2):
\((A+B)+(A-B)=60^\circ+30^\circ\)
\(2A = 90^\circ\)
\(A = 45^\circ\)
Now substitute \(A=45^\circ\) in (1):
\(45^\circ + B = 60^\circ\Rightarrow B=15^\circ\).
Final Answer: \(A=45^\circ\) and \(B=15^\circ\).
Quick Check:
\(A+B=45^\circ+15^\circ=60^\circ\Rightarrow \tan(A+B)=\tan 60^\circ=\sqrt{3}\) ✓
\(A-B=45^\circ-15^\circ=30^\circ\Rightarrow \tan(A-B)=\tan 30^\circ=\dfrac{1}{\sqrt{3}}\) ✓