3. If \( 3 \tan \theta-4=0 \) and \( \cos \theta<0 \) calculate without the use of a cal and with the aid of a diagram the value of \( \begin{array}{ll}\text { (a) } \frac{\sin \theta}{\cos \theta} & \text { (b) } 10 \sin \theta-25 \cos ^{2} \theta\end{array} \) If \( \sin A=\frac{2 \sqrt{6}}{5} \) and \( A \in\left[90^{\circ} ; 360^{\circ}\right] \) calculate without the use of a calculator and with the aid of a diagram the value of 15

To solve \(3 \tan \theta - 4 = 0\), we rearrange it to find \(\tan \theta = \frac{4}{3}\). Since \(\cos \theta < 0\), we identify \(\theta\) is in the second quadrant. Thus, \(\sin \theta = \frac{4}{5}\) and \(\cos \theta = -\frac{3}{5}\). (a) Therefore, \( \frac{\sin \theta}{\cos \theta} = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3} \). (b) Substituting into \(10 \sin \theta - 25 \cos^{2} \theta\): \[ 10 \left(\frac{4}{5}\right) - 25 \left(-\frac{3}{5}\right)^{2} = 8 - 25 \cdot \frac{9}{25} = 8 - 9 = -1. \] Next, for the sine of angle \(A\), given that \(\sin A = \frac{2\sqrt{6}}{5}\) and acknowledging that \(A\) can be either in the second quadrant (where \(\sin A\) is positive and \(\cos A\) is negative) or the fourth (where both sine and cosine are negative), we can proceed to find \(\cos A\) using \(\cos^2 A + \sin^2 A = 1\). That yields \(\cos A = -\frac{\sqrt{17}}{5}\) for the second quadrant. Using the aforementioned properties, we could apply this to find related values or characteristics like \(A\) in degrees!