1. A ladder is leaning against a wall. The base of the la is 4 meters away from the wall, and the ladder reache height of 9 meters on the wall. What is the angle of elevation of the ladder? 3. A person's lines of sight is 1200 meters away from \( t \) hot air balloon flying at 500 meters from the ground. the angle of depression from the hot air balloon to the person?

**Problem 1** 1. Recognize that the ladder, wall, and ground form a right triangle. The vertical side (height on the wall) is \(9\) meters and the horizontal side (distance from the wall) is \(4\) meters. 2. The angle of elevation, \(\theta\), at the base of the ladder can be determined by: \[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{9}{4} \] 3. Solve for \(\theta\): \[ \theta = \tan^{-1} \left(\frac{9}{4}\right) \] 4. Using a calculator, we find: \[ \theta \approx 66^\circ \] --- **Problem 3** 1. In this scenario, the hot air balloon is at a height of \(500\) meters. A person on the ground is horizontally \(1200\) meters away from a point directly beneath the balloon. 2. The angle of depression, \(\theta\), from the balloon to the person is determined by the relation: \[ \tan \theta = \frac{\text{vertical drop}}{\text{horizontal distance}} = \frac{500}{1200} \] 3. Solve for \(\theta\): \[ \theta = \tan^{-1} \left(\frac{500}{1200}\right) \] 4. Calculating this value gives: \[ \theta \approx 22.62^\circ \]