As \(\cos A = \frac{\cos a}{\sin B}\), \(\cos a = \cos A \cdot \frac{\sin B}{\cos B}\). Since \(\frac{\sin B}{\cos B} = \tan B\) (by definition of tangent), we have \(\cos a = \cos A \cdot \tan B\).
Similarly, as \(\sin A = \cos a \cdot \cos B\), \(\cos a = \frac{\sin A}{\cos B}\). Thus \(\sin A = \cos B \cdot \cos a\).
From these two equations we have
\[\cos A \cdot \tan B = \frac{\sin A}{\cos B}.\]
Dividing both sides by \(\frac{1}{\sin A \cos B}\) (which is not zero since \(\sin A \neq 0\) and \(\cos B \neq 0\)) gives us
\[\sin A \cos B = \frac{1}{\frac{\cos A}{\sin B}}.\]
Therefore \[\sin A \cos B = \csc a,\] or \(a = \sin^{-1}\left(\frac{1}{\sin A \cos B}\right)\) to the nearest degree and minute.
Determine the measure of side a ( to the nearest degree and minute).
sin a = sin A cos B
sin A = cos a cos B
cos a = cos B/ sin A
cos a = cos A/ sin B
1 answer
1 answer