4. The angle of depression from the top of a building to a point on the ground is \( 32^{\circ} 30^{\circ} \). How far is the point on the ground from the top of the butlivgs 1 fine brating in 28 high? 5. Find all values of \( \theta \), if \( \theta \) is in the interval \( \left(0^{\circ}, 360^{\circ}\right) \) and has the following function value:

To solve the problems you've presented, let's break them down one by one. ### Problem 4: Angle of Depression The angle of depression from the top of a building to a point on the ground is given as \( 32^\circ 30' \). First, we need to convert this angle into decimal degrees: \[ 32^\circ 30' = 32 + \frac{30}{60} = 32.5^\circ \] Now, if the height of the building is \( h = 28 \) meters, we can use the tangent function to find the horizontal distance \( d \) from the point on the ground to the base of the building. The angle of depression is equal to the angle of elevation from the point on the ground to the top of the building. Using the tangent function: \[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d} \] Substituting the known values: \[ \tan(32.5^\circ) = \frac{28}{d} \] Now, we can solve for \( d \): \[ d = \frac{28}{\tan(32.5^\circ)} \] Calculating \( \tan(32.5^\circ) \): \[ \tan(32.5^\circ) \approx 0.6349 \] Now substituting this value back into the equation for \( d \): \[ d \approx \frac{28}{0.6349} \approx 44.1 \text{ meters} \] So, the point on the ground is approximately **44.1 meters** from the base of the building. ### Problem 5: Finding Values of \( \theta \) You mentioned finding all values of \( \theta \) in the interval \( (0^\circ, 360^\circ) \) for a certain function value, but you did not specify the function or the value. If you provide the specific function (e.g., \( \sin(\theta) = k \), \( \cos(\theta) = k \), \( \tan(\theta) = k \), etc.) and the value \( k \), I can help you find all the angles \( \theta \) that satisfy that equation within the specified interval. Please provide that information for further assistance!