\(A\) and \(B\) are positive angles with \(A>B\) and \(A+B\le 90^\circ\). If \(\cos(A+B)=0\) and \(\cot(A-B)=\displaystyle\frac{1}{\sqrt{3}}\), what is \(B\)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(15^\circ\)
Each condition names a standard angle: in the range \(0^\circ\) to \(90^\circ\), only \(90^\circ\) has cosine 0 and only \(60^\circ\) has cotangent \(\frac{1}{\sqrt{3}}\). Both \(A+B\) and \(A-B\) lie in that range, so they are fixed, and the two equations give \(A\) and \(B\).
\(\cos(A+B)=0\), so \(A+B=90^\circ\)
\(\cot(A-B)=\displaystyle\frac{1}{\sqrt{3}}\), so \(A-B=60^\circ\)
Subtracting: \(2B=90^\circ-60^\circ=30^\circ\), so \(B=15^\circ\) (and \(A=75^\circ\))