I also had a tough time interpreting the question, it’s kinda weird. But I think I’ve figured it out.

It turns out that there is not one diagram that will model this whole question, but rather you’ll have to draw out a few diagrams.

**STEP 1**: Find the distance from the building to firetruck 1 (call this distance x_1). And find the distance from the building to firetruck 2 (call this distance x_2).

So “angle of depression” means, from the people at the top of the building’s perspective, looking down 15^\circ at firetruck 1, and looking down 22^\circ at firetruck 2. So this means that the angle between firetruck 1 and the top of the building is (90^\circ - 15^\circ) = 75^\circ, and the angle between firetruck 2 and the top of the building is (90^\circ - 22^\circ) = 68^\circ.

Now, since we know that the height of the building is 40 meters, we can use the tangent function to find the horizontal distances x_1 and x_2:

\tan (75^\circ) = \dfrac{x_1}{40} \implies x_1 = 40\tan (75^\circ) \approx 149.28\tag*{}

\tan (68^\circ) = \dfrac{x_2}{40} \implies x_2 = 40\tan (68^\circ) \approx 99\tag*{}

**STEP 2:**

Do you know what the law of cosines is? This allows you to calculate the third side of a triangle given that you know what two sides are and the angle between them. In our case, we have calculated x_1 and x_2, and we know that the horizontal angle between them is 32^\circ. This gives us enough information to calculate the distance d between the two trucks, using the law of cosines:

\begin{align*} d &= \sqrt{(x_1)^2 + (x_2)^2 - 2(x_1)(x_2)\cos (32^\circ)} \\ &= \sqrt{(149.28)^2 + (99)^2 - 2(149.28)(99)\cos (32^\circ)} \\ &\approx 83.78 \end{align*}\tag*{}

Here are my diagrams and work: