Building A is 464 feet tall and Building B is 321 feet tall. Dale is standing between the buildings. The angle of elevation from the point on the ground where Dale is standing is 75 degrees to the top of Building A and 48 degrees to the top of Building B, how far apart are the buildings?

The distance between the buildings is:

D = 413.3 ft

Step-by-step explanation:

Ok, here we have two triangle rectangles, such that these triangle rectangles share a common vertex which is Dale.

One of the triangles, let's call it triangle A, has the catheti:

Distance between Dale and Building A = d1

Height of Building A = 464 ft

The other triangle (triangle B) has the catheti:

Distance between Dale and Building B = d2

Height of building B = 321ft

We want to find the distance between the buildings, which is equal to d1 + d2.

Then we want to find d1 and d2.

For triangle A:

We know that the angle at the vertex where Dale is standing, is 75°

Then the adjacent cathetus is d1, and the opposite cathetus is the height of building A.

Remember that:

Tan(θ) = (opposite cathetus)/(adjacent cathetus)

Then:

Tan(75°) = (464 ft)/(d1)

d1 = (464 ft)/(Tan(75°)) = 124.3 ft

Now for triangle B, the angle at the vertex where Dale is standing is 48°.

from this vertex, we have d2 as the adjacent cathetus and the height of build B as the opposite cathetus.

Then:

Tan(48°) = 321ft/d2

d2 = (321ft)/(Tan(48°)) = 289ft

Then the distance between the buildings is:

D = d1 + d2 =  124.3 ft +  289ft = 413.3 ft

hi

step-by-step explanation:

took the test it's 94