A surveyor sights the top of a building with a handheld range finder. The top of the building is 72 feet 10 inches away. The angle of elevation is 57º. Find the distance from the surveyor to the building to the nearest foot.

wat r the choices

A-36

B-38
C-39
D-40

id say 38 wat about u?

How did you get 38? :)

72 ft, 10 inches = 72.8333... ft

cos 57 = base/72.8333..
base = 72.8333..(cos 57°) = appr 39.668 ft

so I would pick D, 40 ft

To find the distance from the surveyor to the building, we can use trigonometry. We'll need to use the tangent function since we have the angle of elevation.

Let's denote the distance from the surveyor to the building as "d". We also know that the height from the ground to the top of the building is 72 feet 10 inches.

Step 1: Convert the height of the building to decimal feet.
Since 1 foot equals 12 inches, we can convert the height to decimal feet by dividing 10 inches by 12:
10 inches / 12 = 0.8333 feet.

Therefore, the height of the building is 72 feet + 0.8333 feet = 72.8333 feet.

Step 2: Find the value of the tangent of the angle of elevation.
The tangent function is defined as the opposite side divided by the adjacent side. In this case, the opposite side is the height of the building, and the adjacent side is the distance from the surveyor to the building (d).

So we can write the equation: tangent(57º) = 72.8333 / d.

Step 3: Solve for 'd'.
Rearranging the equation, we have: d = 72.8333 / tangent(57º).

Using a scientific calculator, evaluate tangent(57º) to find its value.

tangent(57º) ≈ 1.5403.

Now substitute this value back into the equation:
d = 72.8333 / 1.5403.

Calculating this, we find:
d ≈ 47.2693 feet.

Finally, rounding to the nearest foot, the distance from the surveyor to the building is approximately 47 feet.