Given: \( \sin \alpha=\frac{3}{5} \) and \( 90^{\circ}<\alpha<270^{\circ} \) With the aid of a sketch and without using a calculater, detierminve (a) \( \tan \alpha \) (b) \( \cos \left(90^{\circ}+\alpha\right) \)

To start, let's sketch the angle \(\alpha\) knowing that \(90^\circ < \alpha < 270^\circ\). This means \(\alpha\) is in the second or third quadrant. The value given, \(\sin \alpha = \frac{3}{5}\), tells us that in the unit circle representation, the opposite side is 3 and the hypotenuse is 5. Using the Pythagorean theorem to find the adjacent side: \[ \text{adjacent} = \sqrt{\text{hypotenuse}^2 - \text{opposite}^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4. \] Since \(\alpha\) is in the second or third quadrant, where cosine is negative, we take the adjacent side as \(-4\). Now, we can find \(\tan \alpha\): \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}. \] Next, we calculate \(\cos(90^\circ + \alpha)\). By using the trigonometric identity: \[ \cos(90^\circ + \theta) = -\sin \theta, \] we find: \[ \cos(90^\circ + \alpha) = -\sin \alpha = -\frac{3}{5}. \] Thus, the answers are: (a) \( \tan \alpha = -\frac{3}{4} \) (b) \( \cos(90^\circ + \alpha) = -\frac{3}{5} \)