The problem describes an emergency sprinkler projecting water across a 90° angle and is mounted on the roof at a height of 6 m, with the corridor beneath sloping at 13.8°. You are asked to find the length of the ground sprayed by the sprinkler.

Let's define the variables:

• $h = 6 \, \text{m}$: Height of the sprinkler above the ground.
• $\alpha = 13.8^\circ$: Slope angle of the corridor.
• $\theta = \frac{90^\circ}{2} = 45^\circ$: Half the angle of the water spray (since the total angle is divided into two equal parts as per the tip).

We need to calculate the distance along the ground from point $C$ to $B$, where the water touches the ground, given the slope of the ground and the height of the sprinkler.

The key triangle:

You can apply trigonometry to the triangle formed by the sprinkler and the sloped ground. From the diagram:

• The height $h$ forms one side of the triangle.
• The horizontal distance along the ground forms another side.

We need to find the length of $CB$, the distance along the sloped ground. Using trigonometry in the right triangle formed: $\text{Length of } CB = \frac{h}{\sin(\theta + \alpha)}$ Substituting the known values: $CB = \frac{6 \, \text{m}}{\sin(45^\circ + 13.8^\circ)}$ Now let's calculate that value.The length of the ground sprayed by the sprinkler is approximately $7.01 \, \text{m}$, rounded to two decimal places.

Would you like further details on how this was calculated or clarification on any steps?

Here are 5 related questions to consider:

1. How does changing the sprinkler's height affect the sprayed ground distance?
2. What would happen if the slope angle increased to 20°?
3. How can the formula be modified if the sprinkler is moved to a lower height?
4. How does the angle of projection affect the distance covered?
5. What is the effect of the angle between the sprinkler’s water streams on the total ground covered?

Tip: Always double-check trigonometric calculations when working with angles, especially when dealing with slopes and inclined surfaces.